  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 $$\displaystyle \frac{x^{2}}{8}+\frac{y^{2}}{18}=1$$ be minimum, then find the co-ordinates of P.

A
(2,32)  B
(2,3).  C
(2,3).  D
(3,2).  Solution

The correct option is B $$\left ( 2, 3 \right ).$$Given equation of ellipse $$\displaystyle \frac{x^{2}}{8}+\frac{y^{2}}{18}=1$$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 $$\displaystyle \frac{x}{a}\cos t+\frac{y}{b}\sin t= 1$$Its intercepts on the axes are $$\displaystyle \frac{a}{\cos t}$$ and $$\displaystyle \frac{b}{\sin t}$$$$\displaystyle \therefore Area\: of \Delta =\frac{1}{2}\frac{ab}{\sin t\cos t}=\frac{ab}{\sin 2t}$$Area will be least if $$\sin 2t$$ is maximum i.e. $$\displaystyle 2t=90^{\circ}$$ or $$\displaystyle t= 45^{\circ}$$ or $$\displaystyle t= 135^{\circ}$$ (rejected ) as the point P lies in 1st quadrant.$$\displaystyle \therefore P\: is \left ( \frac{a}{\sqrt{2}},\frac{a}{\sqrt{2}} \right )= \left ( \frac{2\sqrt{2}}{\sqrt{2}}, \frac{3\sqrt{2}}{\sqrt{2}} \right )or \left ( 2, 3 \right ).$$Mathematics

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