Maths
General
Easy
Question
If a tangent to ellipse + = 1 makes an angle with x axis, then square of length of intercept of tangent cut between axes is



 None of these
The correct answer is:
Let tangent is += 1
Slope is tan = – S… (1)
square of length of intercept is
=
Length is … (2)
Now use value of tan from (1)
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Vibrating tuning fork of frequency 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 the intensity of sound changes from a maximum to minimum. If the speed of sound is then is
When the piston is moved through a distance of the path difference produced is
This must be equal to for maximum to change to minimum.
So,
This must be equal to for maximum to change to minimum.
So,
Vibrating tuning fork of frequency 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 the intensity of sound changes from a maximum to minimum. If the speed of sound is then is
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When the piston is moved through a distance of the path difference produced is
This must be equal to for maximum to change to minimum.
So,
This must be equal to for maximum to change to minimum.
So,
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PQ and QR are two focal chords of an ellipse and the eccentric angles of P,Q,R and
, respectively then tan tan is equal to 
P (a cos 2 , b sin 2 ), Q (a cos 2 , b sin 2 )
R (a cos 2 , b sin 2 )
chord's PQ equation
cos ( + ) + sin ( + ) = cos ( – )
PQ passes through the focus (ae, 0)
e =
PR passes through the focus (– ae, 0) the
– e =
= –
Apply componendo and dividendo, we get
=
=
tan tan = cot^{2}
PQ and QR are two focal chords of an ellipse and the eccentric angles of P,Q,R and
, respectively then tan tan is equal to 
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P (a cos 2 , b sin 2 ), Q (a cos 2 , b sin 2 )
R (a cos 2 , b sin 2 )
chord's PQ equation
cos ( + ) + sin ( + ) = cos ( – )
PQ passes through the focus (ae, 0)
e =
PR passes through the focus (– ae, 0) the
– e =
= –
Apply componendo and dividendo, we get
=
=
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The radius of the circle passing through the points of intersection of ellipse = 1 and x^{2 }– y^{2} = 0 is 
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Solving = 0 and = 1, we get
=
Therefore radius of circle
= =
Solving = 0 and = 1, we get
=
Therefore radius of circle
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The radius of the circle passing through the points of intersection of ellipse = 1 and x^{2 }– y^{2} = 0 is 
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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.
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=
Therefore radius of circle
= =
Solving = 0 and = 1, we get
=
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= =
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are collinear.
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