Botany-
If there are exactly two points on the ellipse
whose distance from its centre is same and is equal to
then eccentricity of the ellipse is
Botany-General
Answer:The correct answer is:


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Tangent at any point
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is drawn. Eccentric
angle of ‘ P’ is
If
is the foot of perpendicular from centre
to this tangent then
is


Tangent at any point
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If
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botany-General


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An ellipse slides between two perpendicular straight lines
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If the curve
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,
then a possible value of
is
The generated curve is
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should lie inside the circle and out side the ellipse i.e. 
If the curve
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,
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The generated curve is
, whose director circle is
. For the required condition
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Let
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is 16. Then the maximum area of
is
botany-General
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botany-
are points on the ellipse
such that
is a chord through the point
If
then length of chord
is
Conceptual
are points on the ellipse
such that
is a chord through the point
If
then length of chord
is
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If
is a decreasing function for all
and
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represents an ellipse whose major axis is the X-axis is
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If
is a decreasing function for all
and
then the range of K so that the equation
represents an ellipse whose major axis is the X-axis is
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From the focus
of the ellipse
a ray of light is sent which makes angle
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Foci are
Equation of line through (-5, 0) with slope –2 is
This line meets the ellipse above X-axis at
Slope =
.
Foci are
Equation of line through (-5, 0) with slope –2 is
This line meets the ellipse above X-axis at
From the focus
of the ellipse
a ray of light is sent which makes angle
with the positive direction of X-axis upon reacting the ellipse the ray is reflected from it. Slope of the reflected ray is
botany-General
Let 
Foci are
Equation of line through (-5, 0) with slope –2 is
This line meets the ellipse above X-axis at
Slope =
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botany-
If the tangent at Point P to the ellipse 16x2 + 11y2 = 256 is also the tangent to the circle x2 + y2 - 2x = 15, then the eccentric angle of point P is
The equation of tangent at point
to the ellipse
16x2 + 11y2 = 256 is
16x (4cosq) + 11y
= 256
4xcosq +
y sinq = 16
This touches the circle
(x + 1)2 + y2 = 16
So,
= 4
Þ (cosq - 4)2 = 11 + 5cos2q
4cos2q + 8cosq - 5 = 0
\ cosq =
\ q = ±
16x2 + 11y2 = 256 is
16x (4cosq) + 11y
4xcosq +
This touches the circle
(x + 1)2 + y2 = 16
So,
Þ (cosq - 4)2 = 11 + 5cos2q
4cos2q + 8cosq - 5 = 0
\ cosq =
\ q = ±
If the tangent at Point P to the ellipse 16x2 + 11y2 = 256 is also the tangent to the circle x2 + y2 - 2x = 15, then the eccentric angle of point P is
botany-General
The equation of tangent at point
to the ellipse
16x2 + 11y2 = 256 is
16x (4cosq) + 11y
= 256
4xcosq +
y sinq = 16
This touches the circle
(x + 1)2 + y2 = 16
So,
= 4
Þ (cosq - 4)2 = 11 + 5cos2q
4cos2q + 8cosq - 5 = 0
\ cosq =
\ q = ±
16x2 + 11y2 = 256 is
16x (4cosq) + 11y
4xcosq +
This touches the circle
(x + 1)2 + y2 = 16
So,
Þ (cosq - 4)2 = 11 + 5cos2q
4cos2q + 8cosq - 5 = 0
\ cosq =
\ q = ±
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.
If the curve x2 + 3y2 = 9 subtends an obtuse angle at the point (2a, a), then a possible value of a2 is
botany-General
The given curve is
, whose director circle is x2 + y2 = 12. For the required condition (2a, a) should lie inside the circle and outside the ellipse i.e.,(2a)2 + 3a2 - 9 > 0 and (2a)2 + a2 - 12 < 0 i.e.,
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