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Question

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


A
(3,2)
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B
(-2,3)
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C
(2,-3)
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D
(2,3)
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Solution

The correct option is D (2,3)
Let point be $$(2\sqrt 2 sin\theta, 3\sqrt 2 cos\theta)$$
$$y'=\dfrac {-3\sqrt 2}{2\sqrt 2}tan\theta$$
$$y'=\dfrac {-3}{2}tan\theta$$
$$(y-3\sqrt 2cos\theta)=\dfrac {-3}{2}tan\theta (x-2\sqrt 2 sin\theta)$$
$$x=\dfrac {2}{sin\theta}$$ (x intercept)
$$y=\dfrac {3}{cos\theta}$$ (y intercept)
so area $$=\dfrac {1}{2}\dfrac {\times 2\times 3}{sin\theta cos\theta}$$
area is min when $$sin 2\theta=1$$
$$\theta=\dfrac {\pi}{4}$$
so point (2,3)


Mathematics

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