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

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

Answer 1

Equation of the tangent at $\left( \sqrt{24}\sec\theta,\sqrt{18}\tan\theta \right)$ is

$\frac{x\sec\theta}{\sqrt{24}} - \frac{y\tan\theta}{\sqrt{18}} = 1$ .

Since the point of contact is nearest to the line 3x + 2y + 1 = 0,

its slope = – $\frac{3}{2}$ ⇒ $\frac{\sec\theta}{\sqrt{24}}.\frac{\sqrt{18}}{\tan\theta} = - \frac{3}{2}$ ⇒ sinθ = – $\frac{1}{\sqrt{3}}$ .

Hence the point is (6, –3).

Question: The point on the hyperbola \(\frac{x^{2}}{24} - \frac{y^{2}}{18} = 1\) which is nearest to the line ...