Maths-
General
Easy

Question

If P(theta), Q open parentheses theta plus fraction numerator pi over denominator 2 end fraction close parentheses are points on ellipse and alpha is angle between normals at P and Q then -

  1. 2 square root of 1 – e to the power of 2 end exponent end root = text e  end text s i n to the power of 2 end exponent invisible function application 2 theta t a n invisible function application alpha    
  2. 2 square root of 1 – e to the power of 2 end exponent end root = text  e  end text s i n to the power of 2 end exponent invisible function application theta times t a n invisible function application 2 alpha    
  3. square root of 1 – e to the power of 2 end exponent end root= 2 e to the power of 2 end exponent s i n to the power of 2 end exponent invisible function application 2 theta times t a n invisible function application alpha    
  4. 2 square root of 1 – e to the power of 2 end exponent end root = e to the power of 2 end exponent s i n to the power of 2 end exponent invisible function application 2 theta times t a n invisible function application alpha    

The correct answer is: 2 square root of 1 – e to the power of 2 end exponent end root = e to the power of 2 end exponent s i n to the power of 2 end exponent invisible function application 2 theta times t a n invisible function application alpha

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Related Questions to study

General
physics-

The Maxwell's four equations are written as:
i) not stretchy contour integral stack E with rightwards arrow on top. stack d s with rightwards arrow on top equals fraction numerator q subscript 0 end subscript over denominator epsilon subscript 0 end subscript end fraction
ii) not stretchy contour integral stack B with rightwards arrow on top. stack d s with rightwards arrow on top equals 0
iii) not stretchy contour integral stack E with rightwards arrow on top. stack d l with rightwards arrow on top equals fraction numerator d over denominator d t end fraction not stretchy contour integral stack B with rightwards arrow on top. stack d s with rightwards arrow on top
iv) not stretchy contour integral stack B with rightwards arrow on top. stack d s with rightwards arrow on top equals mu subscript 0 end subscript epsilon subscript 0 end subscript fraction numerator d over denominator d t end fraction not stretchy contour integral stack E with rightwards arrow on top. stack d s with rightwards arrow on top
The equations which have sources of stack E with rightwards arrow on top and stack B with rightwards arrow on top

The Maxwell's four equations are written as:
i) not stretchy contour integral stack E with rightwards arrow on top. stack d s with rightwards arrow on top equals fraction numerator q subscript 0 end subscript over denominator epsilon subscript 0 end subscript end fraction
ii) not stretchy contour integral stack B with rightwards arrow on top. stack d s with rightwards arrow on top equals 0
iii) not stretchy contour integral stack E with rightwards arrow on top. stack d l with rightwards arrow on top equals fraction numerator d over denominator d t end fraction not stretchy contour integral stack B with rightwards arrow on top. stack d s with rightwards arrow on top
iv) not stretchy contour integral stack B with rightwards arrow on top. stack d s with rightwards arrow on top equals mu subscript 0 end subscript epsilon subscript 0 end subscript fraction numerator d over denominator d t end fraction not stretchy contour integral stack E with rightwards arrow on top. stack d s with rightwards arrow on top
The equations which have sources of stack E with rightwards arrow on top and stack B with rightwards arrow on top

physics-General
General
maths-

The line x cos alpha + y sin alpha = p touches the ellipse fraction numerator x to the power of 2 end exponent over denominator a to the power of 2 end exponent end fraction plus fraction numerator y to the power of 2 end exponent over denominator b to the power of 2 end exponent end fraction equals 1, if :

The line x cos alpha + y sin alpha = p touches the ellipse fraction numerator x to the power of 2 end exponent over denominator a to the power of 2 end exponent end fraction plus fraction numerator y to the power of 2 end exponent over denominator b to the power of 2 end exponent end fraction equals 1, if :

maths-General
General
maths-

If alpha and beta are the eccentric angles of extremities of a focal chord of an ellipse, then the eccentricity of the ellipse is -

Equation of chord be
fraction numerator x over denominator a end fractioncos fraction numerator alpha plus beta over denominator 2 end fraction + fraction numerator y over denominator b end fractionsinfraction numerator alpha plus beta over denominator 2 end fraction = cosfraction numerator alpha – beta over denominator 2 end fraction
since it passes through (ae, 0)
so e cos fraction numerator alpha plus beta over denominator 2 end fraction = cosfraction numerator alpha – beta over denominator 2 end fraction
e = fraction numerator cos invisible function application fraction numerator alpha – beta over denominator 2 end fraction over denominator cos invisible function application fraction numerator alpha plus beta over denominator 2 end fraction end fraction = fraction numerator 2 cos invisible function application fraction numerator alpha – beta over denominator 2 end fraction sin invisible function application fraction numerator alpha plus beta over denominator 2 end fraction over denominator sin invisible function application left parenthesis alpha plus beta right parenthesis end fraction
e = fraction numerator sin invisible function application alpha plus sin invisible function application beta over denominator sin invisible function application left parenthesis alpha plus beta right parenthesis end fraction

If alpha and beta are the eccentric angles of extremities of a focal chord of an ellipse, then the eccentricity of the ellipse is -

maths-General
Equation of chord be
fraction numerator x over denominator a end fractioncos fraction numerator alpha plus beta over denominator 2 end fraction + fraction numerator y over denominator b end fractionsinfraction numerator alpha plus beta over denominator 2 end fraction = cosfraction numerator alpha – beta over denominator 2 end fraction
since it passes through (ae, 0)
so e cos fraction numerator alpha plus beta over denominator 2 end fraction = cosfraction numerator alpha – beta over denominator 2 end fraction
e = fraction numerator cos invisible function application fraction numerator alpha – beta over denominator 2 end fraction over denominator cos invisible function application fraction numerator alpha plus beta over denominator 2 end fraction end fraction = fraction numerator 2 cos invisible function application fraction numerator alpha – beta over denominator 2 end fraction sin invisible function application fraction numerator alpha plus beta over denominator 2 end fraction over denominator sin invisible function application left parenthesis alpha plus beta right parenthesis end fraction
e = fraction numerator sin invisible function application alpha plus sin invisible function application beta over denominator sin invisible function application left parenthesis alpha plus beta right parenthesis end fraction
General
maths-

If x cos alpha + y sin alpha = p is a tangent to the ellipse fraction numerator x to the power of 2 end exponent over denominator a to the power of 2 end exponent end fraction plus fraction numerator y to the power of 2 end exponent over denominator b to the power of 2 end exponent end fraction equals 1, then -

If x cos alpha + y sin alpha = p is a tangent to the ellipse fraction numerator x to the power of 2 end exponent over denominator a to the power of 2 end exponent end fraction plus fraction numerator y to the power of 2 end exponent over denominator b to the power of 2 end exponent end fraction equals 1, then -

maths-General
General
maths-

Equation of chord of the ellipse fraction numerator x to the power of 2 end exponent over denominator a to the power of 2 end exponent end fraction plus fraction numerator y to the power of 2 end exponent over denominator b to the power of 2 end exponent end fraction equals 1 joining the points P (a cosalpha, b sin alpha) and Q (a cos beta, b sin beta) is 7

Equation of chord of the ellipse fraction numerator x to the power of 2 end exponent over denominator a to the power of 2 end exponent end fraction plus fraction numerator y to the power of 2 end exponent over denominator b to the power of 2 end exponent end fraction equals 1 joining the points P (a cosalpha, b sin alpha) and Q (a cos beta, b sin beta) is 7

maths-General
General
physics-

In Young's double slit experiment, 12 fringes are obtained to be formed in a certain segment of the screen when light of wavelength 600 nm is used. If wavelength of light is changed to 400 nm, number of fringes observed in the same segment of the screen is given by

n subscript 1 end subscript lambda subscript 1 end subscript equals n subscript 2 end subscript lambda subscript 2 end subscript
therefore   n subscript 2 end subscript equals fraction numerator n subscript 1 end subscript lambda subscript 1 end subscript over denominator lambda subscript 2 end subscript end fraction equals fraction numerator 12 x 600 over denominator 400 end fraction equals 18

In Young's double slit experiment, 12 fringes are obtained to be formed in a certain segment of the screen when light of wavelength 600 nm is used. If wavelength of light is changed to 400 nm, number of fringes observed in the same segment of the screen is given by

physics-General
n subscript 1 end subscript lambda subscript 1 end subscript equals n subscript 2 end subscript lambda subscript 2 end subscript
therefore   n subscript 2 end subscript equals fraction numerator n subscript 1 end subscript lambda subscript 1 end subscript over denominator lambda subscript 2 end subscript end fraction equals fraction numerator 12 x 600 over denominator 400 end fraction equals 18
General
physics-

Intensity of central bright fringe due to interference of two identical coherent monochromatic sources is I. If one of the source is switched off, then intensity of central bright fringe becomes

Intensity of central bright fringe due to interference of two identical coherent monochromatic sources is I. If one of the source is switched off, then intensity of central bright fringe becomes

physics-General
General
maths-

If P (a costheta, bsintheta) is a point on an ellipse fraction numerator x to the power of 2 end exponent over denominator a to the power of 2 end exponent end fraction plus fraction numerator y to the power of 2 end exponent over denominator b to the power of 2 end exponent end fraction equals 1, then 'theta ' is –

If P (a costheta, bsintheta) is a point on an ellipse fraction numerator x to the power of 2 end exponent over denominator a to the power of 2 end exponent end fraction plus fraction numerator y to the power of 2 end exponent over denominator b to the power of 2 end exponent end fraction equals 1, then 'theta ' is –

maths-General
General
maths-

If the normal at the point P(theta) to the ellipse fraction numerator x to the power of 2 end exponent over denominator 14 end fraction plus fraction numerator y to the power of 2 end exponent over denominator 5 end fraction equals 1intersects it again at the point Q(2theta) then costheta =

If the normal at the point P(theta) to the ellipse fraction numerator x to the power of 2 end exponent over denominator 14 end fraction plus fraction numerator y to the power of 2 end exponent over denominator 5 end fraction equals 1intersects it again at the point Q(2theta) then costheta =

maths-General
General
maths-

If a tangent to ellipse fraction numerator x to the power of 2 end exponent over denominator a to the power of 2 end exponent end fraction blank+ fraction numerator y to the power of 2 end exponent over denominator b to the power of 2 end exponent end fraction= 1 makes an angle alpha with x- axis, then square of length of intercept of tangent cut between axes is-

Let tangent is fraction numerator x cos invisible function application theta over denominator a end fraction+fraction numerator y sin invisible function application theta over denominator b end fraction= 1
Slope is tan alpha = –fraction numerator b over denominator a end fraction c o t invisible function application theta S… (1)
therefore square of length of intercept is
= left parenthesis a s e c invisible function application theta right parenthesis to the power of 2 end exponent plus left parenthesis b c o s e c invisible function application theta right parenthesis to the power of 2 end exponent
therefore Length is L equals a to the power of 2 end exponent plus b to the power of 2 end exponent plus a to the power of 2 end exponent t a n to the power of 2 end exponent invisible function application theta plus b to the power of 2 end exponent c o t to the power of 2 end exponent invisible function application theta … (2)
Now use value of tan  from (1)

If a tangent to ellipse fraction numerator x to the power of 2 end exponent over denominator a to the power of 2 end exponent end fraction blank+ fraction numerator y to the power of 2 end exponent over denominator b to the power of 2 end exponent end fraction= 1 makes an angle alpha with x- axis, then square of length of intercept of tangent cut between axes is-

maths-General
Let tangent is fraction numerator x cos invisible function application theta over denominator a end fraction+fraction numerator y sin invisible function application theta over denominator b end fraction= 1
Slope is tan alpha = –fraction numerator b over denominator a end fraction c o t invisible function application theta S… (1)
therefore square of length of intercept is
= left parenthesis a s e c invisible function application theta right parenthesis to the power of 2 end exponent plus left parenthesis b c o s e c invisible function application theta right parenthesis to the power of 2 end exponent
therefore Length is L equals a to the power of 2 end exponent plus b to the power of 2 end exponent plus a to the power of 2 end exponent t a n to the power of 2 end exponent invisible function application theta plus b to the power of 2 end exponent c o t to the power of 2 end exponent invisible function application theta … (2)
Now use value of tan  from (1)
General
physics-

Vibrating tuning fork of frequency n is placed near the open end of a long cylindrical tube. The tube has a side opening and is fitted with a movable reflecting piston. As the piston is moved through 8.75 blank c m comma the intensity of sound changes from a maximum to minimum. If the speed of sound is 350 blank m divided by s. then n is

When the piston is moved through a distance of 8.75 c m comma the path difference produced is 2 cross times 8.75 c m equals 17.5 c m.
This must be equal to fraction numerator lambda over denominator 2 end fraction for maximum to change to minimum.
therefore fraction numerator lambda over denominator 2 end fraction equals 17.5 blank c m rightwards double arrow lambda equals 35 c m equals 0.35 m
So, v equals n lambda rightwards double arrow n equals fraction numerator v over denominator lambda end fraction equals fraction numerator 350 over denominator 0.35 end fraction equals 1000 H z

Vibrating tuning fork of frequency n is placed near the open end of a long cylindrical tube. The tube has a side opening and is fitted with a movable reflecting piston. As the piston is moved through 8.75 blank c m comma the intensity of sound changes from a maximum to minimum. If the speed of sound is 350 blank m divided by s. then n is

physics-General
When the piston is moved through a distance of 8.75 c m comma the path difference produced is 2 cross times 8.75 c m equals 17.5 c m.
This must be equal to fraction numerator lambda over denominator 2 end fraction for maximum to change to minimum.
therefore fraction numerator lambda over denominator 2 end fraction equals 17.5 blank c m rightwards double arrow lambda equals 35 c m equals 0.35 m
So, v equals n lambda rightwards double arrow n equals fraction numerator v over denominator lambda end fraction equals fraction numerator 350 over denominator 0.35 end fraction equals 1000 H z
General
maths-

PQ and QR are two focal chords of an ellipse and the eccentric angles of P,Q,R and
2 alpha, 2 beta comma blank 2 gammarespectively then tan beta tan gamma is equal to -

fraction numerator x to the power of 2 end exponent over denominator a to the power of 2 end exponent end fraction plus fraction numerator y to the power of 2 end exponent over denominator b to the power of 2 end exponent end fraction equals 1
P (a cos 2 alpha, b sin 2 alpha), Q (a cos 2 beta , b sin 2 beta)
R (a cos 2 gamma, b sin 2 gamma)
chord's PQ equation
fraction numerator x over denominator a end fractioncos (alpha + beta) + fraction numerator y over denominator b end fractionsin (alpha + beta) = cos (alphabeta)
PQ passes through the focus (ae, 0)
e = fraction numerator cos invisible function application left parenthesis alpha minus beta right parenthesis over denominator cos invisible function application left parenthesis alpha plus beta right parenthesis end fraction
PR passes through the focus (– ae, 0) the
– e = fraction numerator cos invisible function application left parenthesis alpha minus gamma right parenthesis over denominator cos invisible function application left parenthesis alpha plus gamma right parenthesis end fraction
fraction numerator cos invisible function application left parenthesis alpha minus beta right parenthesis over denominator cos invisible function application left parenthesis alpha plus beta right parenthesis end fraction = – fraction numerator cos invisible function application left parenthesis alpha minus gamma right parenthesis over denominator cos invisible function application left parenthesis alpha plus gamma right parenthesis end fraction
Apply componendo and dividendo, we get
fraction numerator cos invisible function application left parenthesis alpha plus beta right parenthesis plus cos invisible function application left parenthesis alpha minus beta right parenthesis over denominator cos invisible function application left parenthesis alpha plus beta right parenthesis minus cos invisible function application left parenthesis alpha minus beta right parenthesis end fraction = fraction numerator cos invisible function application left parenthesis alpha plus gamma right parenthesis minus cos invisible function application left parenthesis alpha minus gamma right parenthesis over denominator cos invisible function application left parenthesis alpha plus gamma right parenthesis plus cos invisible function application left parenthesis alpha minus gamma right parenthesis end fraction
fraction numerator 2 cos invisible function application alpha cos invisible function application beta over denominator 2 sin invisible function application alpha sin invisible function application beta end fraction = fraction numerator 2 sin invisible function application alpha sin invisible function application gamma over denominator 2 cos invisible function application alpha cos invisible function application gamma end fraction
tan beta tan gamma = cot2 alpha

PQ and QR are two focal chords of an ellipse and the eccentric angles of P,Q,R and
2 alpha, 2 beta comma blank 2 gammarespectively then tan beta tan gamma is equal to -

maths-General
fraction numerator x to the power of 2 end exponent over denominator a to the power of 2 end exponent end fraction plus fraction numerator y to the power of 2 end exponent over denominator b to the power of 2 end exponent end fraction equals 1
P (a cos 2 alpha, b sin 2 alpha), Q (a cos 2 beta , b sin 2 beta)
R (a cos 2 gamma, b sin 2 gamma)
chord's PQ equation
fraction numerator x over denominator a end fractioncos (alpha + beta) + fraction numerator y over denominator b end fractionsin (alpha + beta) = cos (alphabeta)
PQ passes through the focus (ae, 0)
e = fraction numerator cos invisible function application left parenthesis alpha minus beta right parenthesis over denominator cos invisible function application left parenthesis alpha plus beta right parenthesis end fraction
PR passes through the focus (– ae, 0) the
– e = fraction numerator cos invisible function application left parenthesis alpha minus gamma right parenthesis over denominator cos invisible function application left parenthesis alpha plus gamma right parenthesis end fraction
fraction numerator cos invisible function application left parenthesis alpha minus beta right parenthesis over denominator cos invisible function application left parenthesis alpha plus beta right parenthesis end fraction = – fraction numerator cos invisible function application left parenthesis alpha minus gamma right parenthesis over denominator cos invisible function application left parenthesis alpha plus gamma right parenthesis end fraction
Apply componendo and dividendo, we get
fraction numerator cos invisible function application left parenthesis alpha plus beta right parenthesis plus cos invisible function application left parenthesis alpha minus beta right parenthesis over denominator cos invisible function application left parenthesis alpha plus beta right parenthesis minus cos invisible function application left parenthesis alpha minus beta right parenthesis end fraction = fraction numerator cos invisible function application left parenthesis alpha plus gamma right parenthesis minus cos invisible function application left parenthesis alpha minus gamma right parenthesis over denominator cos invisible function application left parenthesis alpha plus gamma right parenthesis plus cos invisible function application left parenthesis alpha minus gamma right parenthesis end fraction
fraction numerator 2 cos invisible function application alpha cos invisible function application beta over denominator 2 sin invisible function application alpha sin invisible function application beta end fraction = fraction numerator 2 sin invisible function application alpha sin invisible function application gamma over denominator 2 cos invisible function application alpha cos invisible function application gamma end fraction
tan beta tan gamma = cot2 alpha
General
maths-

The radius of the circle passing through the points of intersection of ellipse fraction numerator x to the power of 2 end exponent over denominator a to the power of 2 end exponent end fraction plus fraction numerator y to the power of 2 end exponent over denominator b to the power of 2 end exponent end fraction = 1 and x2 – y2 = 0 is -

Two curves are symmetrical about both axes and intersect in four points, so, the circle through their points of intersection will have centre at origin.
Solving x to the power of 2 end exponent minus y to the power of 2 end exponent = 0 and fraction numerator x to the power of 2 end exponent over denominator a to the power of 2 end exponent end fraction plus fraction numerator y to the power of 2 end exponent over denominator b to the power of 2 end exponent end fraction= 1, we get
x to the power of 2 end exponent equals y to the power of 2 end exponent = fraction numerator a to the power of 2 end exponent b to the power of 2 end exponent over denominator a to the power of 2 end exponent plus b to the power of 2 end exponent end fraction
Therefore radius of circle
=square root of fraction numerator 2 a to the power of 2 end exponent b to the power of 2 end exponent over denominator a to the power of 2 end exponent plus b to the power of 2 end exponent end fraction end root = fraction numerator square root of 2 a b over denominator square root of a to the power of 2 end exponent plus b to the power of 2 end exponent end root end fraction

The radius of the circle passing through the points of intersection of ellipse fraction numerator x to the power of 2 end exponent over denominator a to the power of 2 end exponent end fraction plus fraction numerator y to the power of 2 end exponent over denominator b to the power of 2 end exponent end fraction = 1 and x2 – y2 = 0 is -

maths-General
Two curves are symmetrical about both axes and intersect in four points, so, the circle through their points of intersection will have centre at origin.
Solving x to the power of 2 end exponent minus y to the power of 2 end exponent = 0 and fraction numerator x to the power of 2 end exponent over denominator a to the power of 2 end exponent end fraction plus fraction numerator y to the power of 2 end exponent over denominator b to the power of 2 end exponent end fraction= 1, we get
x to the power of 2 end exponent equals y to the power of 2 end exponent = fraction numerator a to the power of 2 end exponent b to the power of 2 end exponent over denominator a to the power of 2 end exponent plus b to the power of 2 end exponent end fraction
Therefore radius of circle
=square root of fraction numerator 2 a to the power of 2 end exponent b to the power of 2 end exponent over denominator a to the power of 2 end exponent plus b to the power of 2 end exponent end fraction end root = fraction numerator square root of 2 a b over denominator square root of a to the power of 2 end exponent plus b to the power of 2 end exponent end root end fraction
General
maths-

If  alpha comma beta are the eccentric angles of the extremities of a focal chord of an ellipse, then the eccentricity of the ellipse is -

open parentheses a cos invisible function application alpha comma b sin invisible function application alpha close parentheses comma open parentheses a cos invisible function application beta comma b sin invisible function application beta close parentheses comma open parentheses a e comma 0 close parentheses are collinear.
fraction numerator b left parenthesis sin invisible function application beta – sin invisible function application alpha right parenthesis over denominator a left parenthesis cos invisible function application beta – cos invisible function application alpha right parenthesis end fraction equals fraction numerator b sin invisible function application alpha – 0 over denominator a cos invisible function application alpha – a e end fraction
therefore left parenthesis c o s invisible function application alpha minus e right parenthesis left parenthesis s i n invisible function application beta minus s i n invisible function application alpha right parenthesis equals s i n invisible function application alpha left parenthesis c o s invisible function application beta minus c o s invisible function application alpha right parenthesis
therefore e =blank fraction numerator cos invisible function application alpha left parenthesis sin invisible function application beta – sin invisible function application alpha right parenthesis – sin invisible function application alpha left parenthesis cos invisible function application beta – cos invisible function application alpha right parenthesis over denominator sin invisible function application beta – sin invisible function application alpha end fraction = fraction numerator sin invisible function application left parenthesis alpha – beta right parenthesis over denominator sin invisible function application alpha – sin invisible function application beta end fraction blank= fraction numerator sin invisible function application alpha plus sin invisible function application beta over denominator sin invisible function application left parenthesis alpha plus beta right parenthesis end fraction

If  alpha comma beta are the eccentric angles of the extremities of a focal chord of an ellipse, then the eccentricity of the ellipse is -

maths-General
open parentheses a cos invisible function application alpha comma b sin invisible function application alpha close parentheses comma open parentheses a cos invisible function application beta comma b sin invisible function application beta close parentheses comma open parentheses a e comma 0 close parentheses are collinear.
fraction numerator b left parenthesis sin invisible function application beta – sin invisible function application alpha right parenthesis over denominator a left parenthesis cos invisible function application beta – cos invisible function application alpha right parenthesis end fraction equals fraction numerator b sin invisible function application alpha – 0 over denominator a cos invisible function application alpha – a e end fraction
therefore left parenthesis c o s invisible function application alpha minus e right parenthesis left parenthesis s i n invisible function application beta minus s i n invisible function application alpha right parenthesis equals s i n invisible function application alpha left parenthesis c o s invisible function application beta minus c o s invisible function application alpha right parenthesis
therefore e =blank fraction numerator cos invisible function application alpha left parenthesis sin invisible function application beta – sin invisible function application alpha right parenthesis – sin invisible function application alpha left parenthesis cos invisible function application beta – cos invisible function application alpha right parenthesis over denominator sin invisible function application beta – sin invisible function application alpha end fraction = fraction numerator sin invisible function application left parenthesis alpha – beta right parenthesis over denominator sin invisible function application alpha – sin invisible function application beta end fraction blank= fraction numerator sin invisible function application alpha plus sin invisible function application beta over denominator sin invisible function application left parenthesis alpha plus beta right parenthesis end fraction
General
maths-

If ϕ is the angle between the diameter through any point on a standard ellipse and the normal at the point, then the greatest value of tan ϕ is–

Any point P on ellipse is (a cos theta, b sin theta)
 Equation of the diameter CP is y = open parentheses fraction numerator b over denominator a end fraction tan invisible function application theta close parenthesesx
The normal to ellipse at P is
ax sec theta – by cosec theta = a2e2
Slopes of the lines CP and the normal GP are fraction numerator b over denominator a end fractiontan theta andfraction numerator a over denominator b end fractiontan theta

 tan ϕ = fraction numerator fraction numerator a over denominator b end fraction tan invisible function application theta minus fraction numerator b over denominator a end fraction tan invisible function application theta over denominator 1 plus fraction numerator a over denominator b end fraction tan invisible function application theta. fraction numerator b over denominator a end fraction tan invisible function application theta end fraction=fraction numerator tan invisible function application theta over denominator sec to the power of 2 end exponent invisible function application theta end fraction
=fraction numerator a to the power of 2 end exponent minus b to the power of 2 end exponent over denominator a b end fractionsin  cos  = fraction numerator a to the power of 2 end exponent minus b to the power of 2 end exponent over denominator 2 a b end fractionsin 2
 The greatest value of tan ϕ = fraction numerator a to the power of 2 end exponent minus b to the power of 2 end exponent over denominator 2 a b end fraction.1 = fraction numerator a to the power of 2 end exponent minus b to the power of 2 end exponent over denominator 2 a b end fraction .

If ϕ is the angle between the diameter through any point on a standard ellipse and the normal at the point, then the greatest value of tan ϕ is–

maths-General
Any point P on ellipse is (a cos theta, b sin theta)
 Equation of the diameter CP is y = open parentheses fraction numerator b over denominator a end fraction tan invisible function application theta close parenthesesx
The normal to ellipse at P is
ax sec theta – by cosec theta = a2e2
Slopes of the lines CP and the normal GP are fraction numerator b over denominator a end fractiontan theta andfraction numerator a over denominator b end fractiontan theta

 tan ϕ = fraction numerator fraction numerator a over denominator b end fraction tan invisible function application theta minus fraction numerator b over denominator a end fraction tan invisible function application theta over denominator 1 plus fraction numerator a over denominator b end fraction tan invisible function application theta. fraction numerator b over denominator a end fraction tan invisible function application theta end fraction=fraction numerator tan invisible function application theta over denominator sec to the power of 2 end exponent invisible function application theta end fraction
=fraction numerator a to the power of 2 end exponent minus b to the power of 2 end exponent over denominator a b end fractionsin  cos  = fraction numerator a to the power of 2 end exponent minus b to the power of 2 end exponent over denominator 2 a b end fractionsin 2
 The greatest value of tan ϕ = fraction numerator a to the power of 2 end exponent minus b to the power of 2 end exponent over denominator 2 a b end fraction.1 = fraction numerator a to the power of 2 end exponent minus b to the power of 2 end exponent over denominator 2 a b end fraction .