$\begin{array}{1 1}(A)\;y=\frac{3}{\sqrt 7} x +\frac{15}{\sqrt 7} \\(B)\;y= 3 \sqrt {\frac{2}{7}} x+ \frac{15}{\sqrt 7} \\(C)\;y=2 \sqrt {\frac{3}{7}} x +15 \sqrt 7 \\(D)\;\text{none of these } \end{array}$

Want to ask us a question? Click here

Browse Questions

Ad |

0 votes

0 votes

Let $y=mx+c$ be a common tangent to $9x^2-16y^2=144$ and $x^2+y^2=9$

Since $y=mx+c$ is a tangent to $ 9x^2-16y^2=144$

$\therefore \;c^2=a^2m^2-b^2$

$c^2=16m^2-9$ -----(1)

Now , $y=mx+c$ is a tangent to

$x^2+y^2=9$

$\therefore \large\frac{c}{\sqrt {m^2+1}}$$=3$

=> $ c^2=9(1+m^2)$-----(2)

From (1) and (2) we get

$16m^2-9=9 + 9m^2$

=> $m =\pm 3 \sqrt {\frac{2}{7}}$

Substitute the value of m in eq (2) we get,

$c= \pm \large\frac{15}{\sqrt 7}$

Hence $y= 3 \sqrt {\large\frac{2}{7} }$$x+\large\frac{15}{\sqrt {7}}$ is a common tangent

Hence B 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

...