The angle of intersection of the ellipse frac | vedclass

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(D) To find the intersection,substitute $x^2 = ab - y^2$ into the ellipse equation: $\frac{ab - y^2}{a^2} + \frac{y^2}{b^2} = 1$.This simplifies to $y

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$	an^{-1}\left(\frac{a - b}{\sqrt{ab}}ight)$

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(D) To find the intersection,substitute $x^2 = ab - y^2$ into the ellipse equation: $\frac{ab - y^2}{a^2} + \frac{y^2}{b^2} = 1$.This simplifies to $y^2(\frac{b^2 - a^2}{a^2b^2}) = 1 - \frac{b}{a} = \frac{a - b}{a}$.Solving for $y^2$ and $x^2$,we get $y^2 = \frac{ab^2}{a + b}$ and $x^2 = \frac{a^2b}{a + b}$.The point of intersection is $(x, y) = (a\sqrt{\frac{b}{a+b}}, b\sqrt{\frac{a}{a+b}})$.The slope of the tangent to the ellipse is $m_1 = -\frac{b^2x}{a^2y} = -\frac{b^2}{a^2} \cdot \frac{a}{b} \sqrt{\frac{b}{a}} = -\frac{b}{a} \sqrt{\frac{b}{a}} = -\sqrt{\frac{b^3}{a^3}}$. Wait,simplifying: $m_1 = -\frac{b^2}{a^2} \cdot \frac{a\sqrt{b/(a+b)}}{b\sqrt{a/(a+b)}} = -\frac{b}{a} \sqrt{\frac{b}{a}}$.The slope of the tangent to the circle is $m_2 = -\frac{x}{y} = -\frac{a\sqrt{b/(a+b)}}{b\sqrt{a/(a+b)}} = -\frac{a}{b} \sqrt{\frac{b}{a}} = -\sqrt{\frac{a}{b}}$.The angle $	heta$ is given by $	an 	heta = |\frac{m_1 - m_2}{1 + m_1m_2}|$.Substituting the values: $	an 	heta = |\frac{-\frac{b}{a}\sqrt{\frac{b}{a}} + \sqrt{\frac{a}{b}}}{1 + (\frac{b}{a}\sqrt{\frac{b}{a}})(\sqrt{\frac{a}{b}})}| = |\frac{-\frac{b}{a}\sqrt{\frac{b}{a}} + \sqrt{\frac{a}{b}}}{1 + \frac{b}{a}}| = |\frac{\frac{-b^2 + a^2}{a\sqrt{ab}}}{\frac{a+b}{a}}| = \frac{(a-b)(a+b)}{(a+b)\sqrt{ab}} = \frac{a-b}{\sqrt{ab}}$.Thus,$	heta = 	an^{-1}\left(\frac{a - b}{\sqrt{ab}}ight)$.

The angle of intersection of the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ and the circle $x^2 + y^2 = ab$ is

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