If the minimum area of the triangle formed by a tangent to the ellipse $\frac{x^{2}}{b^{2}}+\frac{y^

Question

If the minimum area of the triangle formed by a tangent to the ellipse $\frac{x^{2}}{b^{2}}+\frac{y^{2}}{4 a^{2}}=1$ and the co-ordinate axis is $k a b$, then $k$ is equal to

Professor PC Thomas Classes · Answer

(2) : Equation of tangent to the given ellipse is, $$\begin{equation*}\frac{x}{b} \cos 	heta+\frac{y}{2 a} \sin 	heta=1 	ag{i} \end{equation*}$$ Now, the tangent (i) cuts $x$-axis and $y$-axis at points $P(b \sec 	heta, 0)$ and $Q(0,2 a \operatorname{cosec} 	heta)$ respectively.\ $	herefore \quad$ Required area $=$ Area of $	riangle P O Q$\ $	herefore A=\frac{1}{2} 	imes b \sec 	heta 	imes 2 a \operatorname{cosec} 	heta$\ $=\frac{2 a b}{2 \sin 	heta \cos 	heta} \Rightarrow A=\frac{2 a b}{\sin 2 	heta}$\ For minimum area, $\sin 2 	heta$ must be maximum and maximum value of $\sin 2 	heta=1$\ $	herefore$ Minimum value of $A=2 a b=k a b$ (Given)\ $\Rightarrow \quad k=2$\