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Standard XII

Mathematics

Position of a Point W.R.T Ellipse

Consider any ...

Question

Consider any point P on the ellipse $\frac{x^{2}}{25} + \frac{y^{2}}{9} = 1$ in the first quadrant. Let $r$ and $s$ represent its distances from (4, 0) and (-4, 0) respectively, then (r+s) is equal to:

10 unit

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9 unit

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8 unit

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6 unit

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Solution

The correct option is A 10 unit
We have, an ellipse $\frac{x^{2}}{25} + \frac{y^{2}}{9} = 1$
Clearly, its foci are $(4, 0)$ and $(- 4, 0)$
$[∵ f o c i = S (a e, 0), S^{'} (- a e, 0)]$
$∴ P S + P S^{'} = 2 a =$ Major Axis
or, $r + s = 2 \times 5 = 10$ units
Hence, A is the correct answer.

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Similar questions

Q. If

P (\frac{π}{4})

is any point on the ellipse

\frac{x^{2}}{25} + \frac{y^{2}}{9} = 1

and

S, S^{'}

are its foci, then find
i) Perimeter of

Δ S P S^{'}

ii)

S P . S^{'} P

Q. On the ellipse

\frac{x^{2}}{8} + \frac{y^{2}}{4} = 1,

let

P

be a point in the second quadrant such that the tangent at

P

to the ellipse is perpendicular to the line

x + 2 y = 0.

Let

S

and

S^{'}

be the foci of the ellipse and

e

be its eccentricity. If

A

is the area of the triangle

S P S^{'},

then the value of

(5 - e^{2}) \cdot A

Let ‘E’ be the ellipse $\frac{x^{2}}{9} + \frac{y^{2}}{4} = 1$ and 'C' be the circle $x^{2} + y^{2} = 9$ . Let P and Q be the points (1, 2) and (2, 1) respectively. Then:

Q. Let

E

be the ellipse

\frac{x^{2}}{9} + \frac{y^{2}}{4} = 1

and

C

be the circle

x^{2} + y^{2} = 9

. Let

P

and

Q

be two points

(1, 2)

and

(2, 1)

respectively. Then:

Q. If

S

and

S^{'}

be the foci of the ellipse

(x^{2} / a^{2}) + (y^{2} / b^{2}) = 1

and

e

be its eccentricity, then

tan \frac{P S S^{'}}{4} tan \frac{P S^{'} S}{4} = \frac{1 - e}{1 + e}

where

P

is any point on the ellipse.

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Consider any point P on the ellipse x225+y29=1 in the first quadrant. Let r and s represent its distances from (4, 0) and (-4, 0) respectively, then (r+s) is equal to:

The correct option is A 10 unitWe have, an ellipse x225+y29=1Clearly, its foci are (4,0) and (−4,0)[∵foci=S(ae,0),S′(−ae,0)]∴PS+PS′=2a=Major Axisor, r+s=2×5=10 units Hence, A is the correct answer.

Consider any point P on the ellipse $\frac{x^{2}}{25} + \frac{y^{2}}{9} = 1$ in the first quadrant. Let $r$ and $s$ represent its distances from (4, 0) and (-4, 0) respectively, then (r+s) is equal to:

The correct option is A 10 unit
We have, an ellipse $\frac{x^{2}}{25} + \frac{y^{2}}{9} = 1$
Clearly, its foci are $(4, 0)$ and $(- 4, 0)$
$[∵ f o c i = S (a e, 0), S^{'} (- a e, 0)]$
$∴ P S + P S^{'} = 2 a =$ Major Axis
or, $r + s = 2 \times 5 = 10$ units
Hence, A is the correct answer.