\[\begin{array}{1 1}(1)\bigg(\frac{3}{8},24\bigg)&(2)\bigg(-24,\frac{-3}{8}\bigg)\\(3)\bigg(\frac{-3}{8},-24\bigg)&(4)\bigg(24,\frac{3}{8}\bigg)\end{array}\]

Want to ask us a question? Click here

Browse Questions

Ad |

0 votes

0 votes

The equation of the given hyperbola is $xy= 9$

$( 6,\large\frac{3}{2})$ is a point on it

$(ct_1,\large\frac{c}{t_1} )$$ =(6,\large\frac{3}{2})$

$ct_1=6,= \large\frac{c}{t_1} =\frac{3}{2}$

$c=3,t_1= 2$

Let the normal at $(ct, \large\frac{c}{t_1})$ meets the curve again at $(ct_2, \large\frac{c}{t_2})$ the $t_1^3t_2=-1$

$t_2= \large\frac{-1}{t_1^3}=\frac{-1}{8}$

The required point is $\bigg(\frac{-3}{8},-24\bigg)$

Hence 3 is the correct answer.

Ask Question

Take Test

x

JEE MAIN, CBSE, NEET Mobile and Tablet App

The ultimate mobile app to help you crack your examinations

...