The minimum area of a triangle formed by any | vedclass

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(D) Let the point of tangency be $(h, k)$ on the ellipse $\frac{x^2}{16} + \frac{y^2}{81} = 1$.The equation of the tangent at $(h, k)$ is given by $\f

Vedclass · Accepted Answer

$36$

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(D) Let the point of tangency be $(h, k)$ on the ellipse $\frac{x^2}{16} + \frac{y^2}{81} = 1$.The equation of the tangent at $(h, k)$ is given by $\frac{xh}{16} + \frac{yk}{81} = 1$.The $x$-intercept is found by setting $y=0$,which gives $x = \frac{16}{h}$. So,point $B = (\frac{16}{h}, 0)$.The $y$-intercept is found by setting $x=0$,which gives $y = \frac{81}{k}$. So,point $A = (0, \frac{81}{k})$.The area of the triangle $OAB$ is $A = \frac{1}{2} 	imes |\frac{16}{h}| 	imes |\frac{81}{k}| = \frac{648}{|hk|}$.Since $(h, k)$ lies on the ellipse,$\frac{h^2}{16} + \frac{k^2}{81} = 1$. By the $AM$-$GM$ inequality,$\frac{\frac{h^2}{16} + \frac{k^2}{81}}{2} \ge \sqrt{\frac{h^2 k^2}{16 	imes 81}}$.$\frac{1}{2} \ge \frac{|hk|}{4 	imes 9} \Rightarrow |hk| \le 18$.Therefore,the minimum area $A = \frac{648}{|hk|} \ge \frac{648}{18} = 36$.The minimum area of the triangle is $36$ square units.

The minimum area of a triangle formed by any tangent to the ellipse $\frac{x^2}{16} + \frac{y^2}{81} = 1$ and the coordinate axes is

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