The tangent at point (a cos theta, b sin thet | vedclass

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(B) 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

Vedclass · Accepted Answer

$2ab$

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(B) 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 $\frac{x \cos 	heta}{a} + \frac{y \sin 	heta}{b} = 1$.To find the $x$-intercept $T$,set $y = 0$: $\frac{x \cos 	heta}{a} = 1 \implies x = \frac{a}{\cos 	heta}$. Thus,$OT = \frac{a}{\cos 	heta}$.To find the $y$-intercept $T_1$,set $x = 0$: $\frac{y \sin 	heta}{b} = 1 \implies y = \frac{b}{\sin 	heta}$. Thus,$OT_1 = \frac{b}{\sin 	heta}$.The product $(OT)(OT_1) = \frac{a}{\cos 	heta} \cdot \frac{b}{\sin 	heta} = \frac{ab}{\sin 	heta \cos 	heta} = \frac{2ab}{\sin 2	heta}$.Since $0 Therefore,the minimum value is $2ab$.

The tangent at point $(a \cos \theta, b \sin \theta)$,where $0 < \theta < \frac{\pi}{2}$,to the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ meets the $x$-axis at $T$ and the $y$-axis at $T_1$. Then the value of $\min_{0 < \theta < \frac{\pi}{2}} (OT)(OT_1)$ is

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