The minimum area of a triangle formed by any | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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

$36$

Vedclass · Answer

(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

Similar Questions

The value of $k$,if $(1, 2)$ and $(k, -1)$ are conjugate points with respect to the ellipse $2x^2 + 3y^2 = 6$,is

If $x+y+n=0, n>0$ is a normal to the ellipse $x^2+3y^2=3$ and $x+my+3=0, m < 0$ is a tangent to the ellipse $x^2+5y^2=5$,then the point of intersection of these two lines satisfies the equation

The distance between the directrices of the conic $\sqrt{(x - 5)^2 + (y - 4)^2} + \sqrt{(x - 3)^2 + (y - 2)^2} = 6$ is

If the latus rectum of an ellipse is equal to half of its minor axis,then its eccentricity is

Vedclass Test Series

Exam Paper Generator

Online Exam Module