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 $S^{\prime}$ be the foci of the ellipse and e be its eccentricity. If $A$ is the area of the triangle SPS' then, the value of $\left(5-e^{2}\right) . A$ is:

JEE Main · 2021 · 26 Aug 2021 · Shift 1 · Ellipse

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 SPS' then, the value of

(5 - e^{2}) . A

is:

Answer: (B)

Here

a^{2} = 8

and

b^{2} = 4

, so

e^{2} = 1 - \frac{4}{8} = \frac{1}{2}

c^{2} = a^{2} - b^{2} = 4

and the foci are

S (2, 0)

and

S^{'} (- 2, 0)

.
Differentiating the ellipse equation gives

\frac{d y}{d x} = - \frac{x}{2 y}

. The tangent at

P

is perpendicular to

x + 2 y = 0

, so its slope is

2

, and

- \frac{x}{2 y} = 2

gives

x = - 4 y

.
Substituting in the ellipse,

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

, so

y^{2} = \frac{4}{9}

; in the second quadrant

P = (- \frac{8}{3}, \frac{2}{3})

.
Taking

S S^{'} = 4

as base, the height of the triangle is the ordinate of

P

, so

A = \frac{1}{2} \cdot 4 \cdot \frac{2}{3} = \frac{4}{3}

and

(5 - e^{2}) A = \frac{9}{2} \cdot \frac{4}{3} = 6.

JEE Main · 2021 · 26 Aug 2021 · Shift 1 · Single correct option