The angle of intersection of ellipse frac{{{x | vedclass

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Question

(d) $\frac{{ab - {y^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} = 1$or ${y^2}\left( {\frac{{{a^2} - {b^2}}}{{{a^2}{b^2}}}} \right) = \frac{{a - b}}{a}$

Vedclass · Accepted Answer

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

Vedclass · Answer

(d) $\frac{{ab - {y^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} = 1$or ${y^2}\left( {\frac{{{a^2} - {b^2}}}{{{a^2}{b^2}}}} ight) = \frac{{a - b}}{a}$

or ${y^2} = \left( {\frac{{a{b^2}}}{{a + b}}} ight)$and ${x^2} = \left( {\frac{{{a^2}b}}{{a + b}}} ight)\,\, $

$\Rightarrow \,(x,\,y) = \left( {a\sqrt {\frac{b}{{a + b}}} ,b\sqrt {\frac{a}{{a + b}}} } ight)$

Slope of tangent at ellipse $ = \frac{{ - {b^2}x}}{{{a^2}y}} = \frac{{ - {b^2}}}{{{a^2}}}\sqrt {\frac{a}{b}} $

Slope of tangent at circle $ = - \frac{x}{y} = - \sqrt {\frac{a}{b}} $

$	herefore 	heta = {	an ^{ - 1}}\left[ {\frac{{ - \sqrt {\frac{a}{b}} + \frac{{{b^2}}}{{{a^2}}}\sqrt {\frac{a}{b}} }}{{1 + \frac{{{b^2}}}{{{a^2}}}.\frac{a}{b}}}} ight]$

$i.e.$, $	heta = {	an ^{ - 1}}\left[ {\frac{{a - b}}{{\sqrt {ab} }}} ight]$.

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

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