Let E_1 and E_2 be two ellipses whose centers | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) For the line $x+y=3$,the point of contact for $E_1: \frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ is $(\frac{a^2}{3}, \frac{b^2}{3})$ and for $E_2: \frac{x^2

Vedclass · Accepted Answer

$(A, B)$

Vedclass · Answer

(A) For the line $x+y=3$,the point of contact for $E_1: \frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ is $(\frac{a^2}{3}, \frac{b^2}{3})$ and for $E_2: \frac{x^2}{B^2}+\frac{y^2}{A^2}=1$ is $(\frac{B^2}{3}, \frac{A^2}{3})$.The point of contact of $x+y=3$ with the circle $x^2+(y-1)^2=2$ is $(1, 2)$.The general point on $x+y=3$ at a distance $r$ from $(1, 2)$ is $(1 \mp \frac{r}{\sqrt{2}}, 2 \pm \frac{r}{\sqrt{2}})$. Given $r=\frac{2 \sqrt{2}}{3}$,the points $Q$ and $R$ are $(\frac{1}{3}, \frac{8}{3})$ and $(\frac{5}{3}, \frac{4}{3})$.Comparing these with the points of contact,we get $a^2=5, b^2=4$ for $E_1$ and $B^2=1, A^2=8$ for $E_2$ (or vice-versa).For $E_1$,$e_1^2 = 1 - \frac{4}{5} = \frac{1}{5}$. For $E_2$,$e_2^2 = 1 - \frac{1}{8} = \frac{7}{8}$.Then $e_1^2+e_2^2 = \frac{1}{5} + \frac{7}{8} = \frac{8+35}{40} = \frac{43}{40}$.And $e_1 e_2 = \sqrt{\frac{1}{5} \cdot \frac{7}{8}} = \sqrt{\frac{7}{40}} = \frac{\sqrt{7}}{2 \sqrt{10}}$.Thus,options $(A)$ and $(B)$ are correct.

Similar Questions

At which point on the ellipse $4x^2 + 9y^2 = 1$ is the tangent parallel to $8x = 9y$?

Let $P$ be a variable point on the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ with foci $F_1$ and $F_2$. If $A$ is the area of triangle $PF_1F_2$,then the maximum value of $A$ is:

If the angle between the lines joining the end points of the minor axis of an ellipse with its foci is $\frac{\pi}{2}$,then the eccentricity of the ellipse is

The length of the latus rectum of $9x^2 + 25y^2 - 90x - 150y + 225 = 0$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module