Test Index

Ellipse

Section: Mathematics

© examsnet.com

Question : 12 of 25

Marks: +1, -0

Let

E_1

E_2

be two ellipse whose centers are at the origin.The major axes of

E_1

E_2

lie along the x-axis and the y-axis,respectively,Let

S

x^2+(y+1)^2=2

The straight line x+y=3 touches the curves

S

E_1

E_2

P,Q

R

,respectively.Suppose that

PQ=PR=\frac{2\sqrt{2}}{2}

e_1

e_2

are the eccentricities of

E_1

E_2

,respectively,then the correct expression(s) is (are)

[JEE Adv 2015 P2]

$e_1^2+e_2^2=\frac{43}{40}$
$e_1e_2=\frac{\sqrt{7}}{2\sqrt{10}}$
$\left|e_1^2-e_2^2\right|=\frac{5}{8}$
$e_1e_2=\frac{\sqrt{3}}{4}$

Solution:

For the given line point of contact for

E_1:\frac{x^2}{a^2}+\frac{y^2}{b^2}=1

is

\left(\frac{a^2}{3},\frac{b^2}{3}\right)

and for

E_2:\frac{x^2}{B^2}+\frac{y^2}{A^2}=1

is

\left(\frac{B^2}{3},\frac{A^2}{3}\right)

Point of contact of

x+y=3

and circle is

(1,2)

Also,general point on

x+y=3

can be taken as

\left(1\mp\frac{r}{\sqrt{2}},2\pm\frac{r}{\sqrt{2}}\right)

where

r=\frac{2\sqrt{2}}{3}

So,required points are

\left(\frac{1}{3},\frac{8}{3}\right)

and

\left(\frac{5}{3},\frac{4}{3}\right)

Comparing with points of contact of ellipse.

a^2=5,B^2=8

b^2=4,A^2=1

∴

e_1e_2=\frac{\sqrt{7}}{2\sqrt{10}}

and

e_1^2+e_2^2=\frac{43}{40}

© examsnet.com

Go to Question:

Prev Question

Next Question