Question

The point on the hyperbola $\frac{x^2}{24}-\frac{y^2}{18}=1$ which is nearest to the line $3x+2y+1=0$ is

• A. $(-6,3)$
• B. $(6,-3)$
• C. $(6,3)$
• D. None of these

Answer 1

The correct option is B $(6,-3)$

The coordinates of any point on the hyperbola are $(\sqrt{24}\sec \theta ,\sqrt{18}\tan \theta ).$

Equation of tangent at this point is $\frac{x\sec \theta }{\sqrt{24}}-\frac{y\tan \theta }{\sqrt{18}}=1$ ...(1)

The point is nearest to the line $3x+2y+1=0$ ...(2)

If (1) and (2) are parallel $\Rightarrow \frac{\sec \theta }{\sqrt{24}}.\frac{\sqrt{18}}{\tan \theta }=-\frac{3}{2}\Rightarrow \sin \theta =-\frac{1}{\sqrt{3}}$

Thus, the point is $(6,-3)$.