The minimum area of the triangle formed by an | vedclass

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(C) The equation of the tangent to the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ at the point $(a \cos 	heta, b \sin 	heta)$ is given by $\fra

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$ab$

Vedclass · Answer

(C) The equation of the tangent to the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ at the point $(a \cos 	heta, b \sin 	heta)$ is given by $\frac{x}{a} \cos 	heta + \frac{y}{b} \sin 	heta = 1$.The $x$-intercept of this tangent is found by setting $y = 0$,which gives $P = (\frac{a}{\cos 	heta}, 0)$.The $y$-intercept of this tangent is found by setting $x = 0$,which gives $Q = (0, \frac{b}{\sin 	heta})$.The area of the triangle $OPQ$ formed by the tangent and the coordinate axes is $A = \frac{1}{2} 	imes |x_P| 	imes |y_Q| = \frac{1}{2} \left| \frac{a}{\cos 	heta} ight| \left| \frac{b}{\sin 	heta} ight| = \frac{ab}{|2 \sin 	heta \cos 	heta|} = \frac{ab}{|\sin 2	heta|}$.Since the minimum value of $|\sin 2	heta|$ is $1$ (when $2	heta = 90^\circ$ or $270^\circ$),the minimum area is $ab$.

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

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