Question

The coordinates of a point P(x, y) lying in the first quadrant of 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 axes is the smallest are

• A. (3,2)
• B. (-2,3)
• C. (2,-3)
• D. (2,3)

Answer 1

The correct option is D (2,3)

Let point be $(2\sqrt{2}sin\theta ,3\sqrt{2}cos\theta )$

$y^{\prime }=\frac{-3\sqrt{2}}{2\sqrt{2}}tan\theta$

$y^{\prime }=\frac{-3}{2}tan\theta$

$(y-3\sqrt{2}cos\theta )=\frac{-3}{2}tan\theta (x-2\sqrt{2}sin\theta )$

$x=\frac{2}{sin\theta }$ (x intercept)

$y=\frac{3}{cos\theta }$ (y intercept)

so area $=\frac{1}{2}\frac{\times 2\times 3}{sin\theta cos\theta }$

area is min when $sin2\theta =1$

$\theta =\frac{\pi }{4}$

so point (2,3)