Consider the curve frac{x^{2}}{a^{2}}+frac{y^ | vedclass

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(C) Let the ellipse be $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$ with $a > b$. Let $P(a \cos 	heta, b \sin 	heta)$ be a point on the ellipse. The

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$\frac{\pi}{2}$

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(C) Let the ellipse be $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$ with $a > b$. Let $P(a \cos 	heta, b \sin 	heta)$ be a point on the ellipse. The equation of the tangent at $P$ is $\frac{x \cos 	heta}{a} + \frac{y \sin 	heta}{b} = 1$. The directrix is $x = \frac{a}{e}$. Substituting $x = \frac{a}{e}$ into the tangent equation: $\frac{(\frac{a}{e}) \cos 	heta}{a} + \frac{y \sin 	heta}{b} = 1 \implies \frac{\cos 	heta}{e} + \frac{y \sin 	heta}{b} = 1$. $y = \frac{b(1 - \frac{\cos 	heta}{e})}{\sin 	heta} = \frac{b(e - \cos 	heta)}{e \sin 	heta}$. Let $S(ae, 0)$ be the focus. The slope of $SP$ is $m_1 = \frac{b \sin 	heta}{a \cos 	heta - ae} = \frac{b \sin 	heta}{a(\cos 	heta - e)}$. The slope of the line segment from the focus to the intersection point $Q(\frac{a}{e}, \frac{b(e - \cos 	heta)}{e \sin 	heta})$ is $m_2 = \frac{\frac{b(e - \cos 	heta)}{e \sin 	heta} - 0}{\frac{a}{e} - ae} = \frac{b(e - \cos 	heta)}{e \sin 	heta} 	imes \frac{e}{a(1 - e^{2})} = \frac{b(e - \cos 	heta)}{a \sin 	heta (1 - e^{2})}$. Since $b^{2} = a^{2}(1 - e^{2})$, we have $1 - e^{2} = \frac{b^{2}}{a^{2}}$. $m_2 = \frac{b(e - \cos 	heta)}{a \sin 	heta (\frac{b^{2}}{a^{2}})} = \frac{a(e - \cos 	heta)}{b \sin 	heta}$. Note that $m_1 	imes m_2 = \frac{b \sin 	heta}{a(\cos 	heta - e)} 	imes \frac{a(e - \cos 	heta)}{b \sin 	heta} = -1$. Therefore, the angle subtended is $\frac{\pi}{2}$.

Consider the curve $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$. The portion of the tangent at any point of the curve intercepted between the point of contact and the directrix subtends at the corresponding focus an angle of

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