Ask Questions, Get Answers

Want to ask us a question? Click here
Browse Questions
0 votes

The product of the perpendiculars drawn from the point $(8 ,0)$ on the hyperbola to its asymptotes is $\large\frac{x^{2}}{64}-\frac{y^{2}}{36}=$$1$ is

\[\begin{array}{1 1}(1)\frac{25}{576}&(2)\frac{576}{25}\\(3)\frac{6}{25}&(4)\frac{25}{6}\end{array}\]

Can you answer this question?

1 Answer

0 votes
The equation of the hyperbola is
$\large\frac{x62}{64} -\frac{y^2}{36}$
$a^2= 64,b^2= 36$
Required length of the perpendicular $=\large\frac{a^2b^2}{a^2+b^2}$
$\qquad= \large\frac{64 \times 36}{64 +36}$
$\qquad= \large\frac{576}{25}$
Hence 2 is the correct answer.
answered May 16, 2014 by meena.p
Ask Question
student study plans
JEE MAIN, CBSE, AIPMT Mobile and Tablet App
The ultimate mobile app to help you crack your examinations
Get the Android App