Search any question & find its solution
Question:
Answered & Verified by Expert
If the product of the lengths of the perpendiculars drawn from the foci to the tangent $y=\frac{-3}{4} x+3 \sqrt{2}$ of the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ is 9 , then the eccentricity of that ellipse is
Options:
Solution:
2442 Upvotes
Verified Answer
The correct answer is:
$\frac{\sqrt{7}}{4}$
We know that, the product of the length of the perpendicular drawn from the foci to the tangent of the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ is $b^2$.
Also, length $=9 \quad \therefore \quad b^2=9$
Equation of tangent
$y=\frac{-3}{4} x+3 \sqrt{2}$
$\therefore \quad 3 \sqrt{2}= \pm \sqrt{a^2\left(\frac{-3}{4}\right)^2+b^2}$
$18=\frac{9}{16} a^2+9 \Rightarrow 2-1=\frac{a^2}{16} \Rightarrow a^2=16$
$\therefore \quad e=\sqrt{1-\frac{b^2}{a^2}}=\sqrt{1-\frac{9}{16}}=\frac{\sqrt{7}}{4}$
Also, length $=9 \quad \therefore \quad b^2=9$
Equation of tangent
$y=\frac{-3}{4} x+3 \sqrt{2}$
$\therefore \quad 3 \sqrt{2}= \pm \sqrt{a^2\left(\frac{-3}{4}\right)^2+b^2}$
$18=\frac{9}{16} a^2+9 \Rightarrow 2-1=\frac{a^2}{16} \Rightarrow a^2=16$
$\therefore \quad e=\sqrt{1-\frac{b^2}{a^2}}=\sqrt{1-\frac{9}{16}}=\frac{\sqrt{7}}{4}$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.