Test Index

Ellipse

Section: Mathematics

© examsnet.com

Question : 72 of 117

Marks: +1, -0

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+2y=0

. Let S and S' be the foci of the ellipse and e be its eccentricity. If A is the area of the

\Delta S P S'

then, the value of

(5-e^{2}).

[26 Aug 2021 Shift 1]

6
12
14
24

Solution:

Given, equation of ellipse

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

Then, equation of tangent at

(x_{1}, y_{1})

will be

\frac{x x_{1}}{8}+\frac{y y_{1}}{4}=1

Since, tangent is perpendicular to the line

x+2y=0

, then

\left(\frac{-x_{1}}{2 y_{1}}\right)\left(\frac{-1}{2}\right)=-1

\Rightarrow x_{1}=-4 y_{1}

Also,

\frac{x_{1}^{2}}{8}+\frac{y_{1}^{2}}{4}=1

\Rightarrow \frac{16 y_{1}^{2}}{8}+\frac{y_{1}^{2}}{4}=1

[\because x_{1}=-4 y_{1}]

\Rightarrow \frac{9}{4} y_{1}^{2}=1

\Rightarrow y_{1}^{2}=\frac{4}{9}

\Rightarrow y_{1}=\pm\frac{2}{3}

\Rightarrow y_{1}=\frac{2}{3}[\because (x_{1}, y_{1}).

lies in second quadrant]

And

x_{1}=-4 y_{1}=-4 \times \frac{2}{3}=\frac{-8}{3}

∴

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

Again,

a^{2}-b^{2}=a^{2}e^{2}

8-4=8e^{2}

e=\frac{1}{\sqrt{2}}

Now, S and S' are the foci of the ellipse,

So, S:(ae, 0)

=\left(2 \sqrt{2} \cdot \frac{1}{\sqrt{2}},0\right)=(2,0)

and

S':(-a e, 0)=\left(-2 \sqrt{2} \cdot \frac{1}{\sqrt{2}},0\right)=(-2,0)

Area of

\Delta SPS'=\frac{1}{2} \times 4 \times \frac{2}{3} = \frac{4}{3} \; [\because

base = 4

, \text{height} = \frac{2}{3}

]

So,

(5-e^{2})\; A = \left(5-\frac{1}{2}\right) \frac{4}{3} = \frac{9}{2} \cdot \frac{4}{3} = 6

© examsnet.com

Go to Question:

Prev Question

Next Question