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

Name: PDF Generator App | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

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

Similar Questions

Find the equation for the ellipse that satisfies the given conditions: Ends of major axis $(±3,\,0)$ ends of minor axis $(0,\,±2)$

An ellipse inscribed in a semi-circle touches the circular arc at two distinct points and also touches the bounding diameter. Its major axis is parallel to the bounding diameter. When the ellipse has the maximum possible area, its eccentricity is

The angle between the pair of tangents drawn from the point $(1, 2)$ to the ellipse $3{x^2} + 2{y^2} = 5$ is

If end points of latus rectum of an ellipse are vertices of a square, then eccentricity of ellipse will be -