Question

If the area of the triangle formed by the axes of co-ordinates and tangent at any point P(x, y) in 1st quadrant on the ellipse $\frac{x^2}{8}+\frac{y^2}{18}=1$ be minimum, then find the co-ordinates of P.

• A. $(2,3\sqrt{2})$
• B. $(\sqrt{2},3).$
• C. $(2,3).$
• D. $(3,2).$

Answer 1

Answer: (C)

$(2,3).$

Given equation of ellipse $\frac{x^2}{8}+\frac{y^2}{18}=1$

Here, $a^2=8,b^2=18$

Let $P(a\cos t,b\sin t)$ be a point on the ellipse

Equation of tangent at $P(a\cos t,b\sin t)$ is

$\frac{x}{a}\cos t+\frac{y}{b}\sin t=1$

Its intercepts on the axes are $\frac{a}{\cos t}$ and $\frac{b}{\sin t}$
$\therefore Areaof\Delta =\frac{1}{2}\frac{ab}{\sin t\cos t}=\frac{ab}{\sin 2t}$

Area will be least if $\sin 2t$ is maximum i.e. $2t={90}^{\circ }$ or $t={45}^{\circ }$ or $t={135}^{\circ }$ (rejected ) as the point P lies in 1st quadrant.
$\therefore Pis(\frac{a}{\sqrt{2}},\frac{a}{\sqrt{2}})=(\frac{2\sqrt{2}}{\sqrt{2}},\frac{3\sqrt{2}}{\sqrt{2}})or(2,3).$