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

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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