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. $(\sqrt{18},0)$
• D. (3,2)

Answer 1

The correct option is A (2, 3)

Any point on the ellipse 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 $(\sqrt{18}cos\theta ,\sqrt{18}sin\theta )$ is
y - $\sqrt{18}sin\theta =-\frac{\sqrt{9}}{2}cot\theta (x-\sqrt{8}cos\theta )$
Therefore, the tangent cuts the coordinates axes at the points $(0,\frac{\sqrt{18}}{sin\theta })$ and $(\frac{\sqrt{18}}{cos\theta },0)$
Thus the area of the triangle formed by this tangent and the coordinate axes is A = $\frac{1}{2}\sqrt{18}\sqrt{8}$ . $\frac{1}{cos\theta sin\theta }=\frac{6}{cos\theta sin\theta }=12cosec2\theta$.
But $cosec2\theta$ is smallest when $\theta =\frac{\pi }{4}$. Therefore A is smallest when $\theta =\frac{\pi }{4}$.
Hence the required point is $(\sqrt{8}.\frac{1}{\sqrt{2}},\sqrt{18}.\frac{1}{\sqrt{2}})$ = (2, 3)