Question

The coordinates of the point $P(x,y)$ lying in the first quadrant on the ellipse $\frac{x^2}{8}+\frac{y^2}{18}=1$ so that the area of the triangle formed by the tangent at $P$ and the coordinate axes is the smallest, are given by

• A. $(2,3)$
• B. $(\sqrt{8},0)$
• C. $(0,\sqrt{18})$
• D. none of these

Answer 1

The correct option is A $(2,3)$

Any point on the ellipse $\frac{x^2}{8}+\frac{y^2}{18}=1$ is given by $(\sqrt{8}\cos \theta ,\sqrt{18}\sin \theta )$

Now, $\frac{2x}{8}+\frac{2}{18}y\frac{dy}{dx}=0\Rightarrow \frac{dy}{dx}=-\frac{9x}{4y}$

$\Rightarrow {\frac{dy}{dx}|}_{(\sqrt{8}\cos \theta ,\sqrt{18}\sin \theta )}=-\frac{9\sqrt{8}\cos \theta }{4\sqrt{18}\sin \theta }$

$=-\frac{\sqrt{9}}{2}\cot \theta$

Hence, the equation of the tangent at $(\sqrt{8}\cos \theta ,\sqrt{18}\sin \theta )$ is

$y-\sqrt{18}\sin \theta =-\frac{\sqrt{9}}{2}\cot \theta (x-\sqrt{8}\cos \theta )$

Thus, the area of the triangle formed by this tangent and the coordinate axes is

$A=\frac{1}{2}\sqrt{18}\cdot \sqrt{8}\cdot \frac{1}{\cos \theta \sin \theta }=\frac{6}{\cos \theta \sin \theta }=12cosec2\theta$

But $cosec2\theta$ is smallest when $\theta =\frac{\pi }{4}$,therefore $A$ is smallest when $\theta =\pi /4$.

Hence the required point is

$(\sqrt{8}\cdot \frac{1}{\sqrt{2}},\sqrt{18}\cdot \frac{1}{\sqrt{2}})\equiv (2,3)$

Ans: A