A point on the ellipse, 4x^2 + 9y^2 = 36, whe | vedclass

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Question

Given ellipse is $4{x^2} + 9{y^2} = 36$

$ \Rightarrow \frac{{{x^2}}}{9} + \frac{{{y^2}}}{4} = 1$

Normal at the point is parallel to the lin

Vedclass · Accepted Answer

$\left( {-\frac{9}{5},\frac{8}{5}} \right)$

Vedclass · Answer

Given ellipse is $4{x^2} + 9{y^2} = 36$

$ \Rightarrow \frac{{{x^2}}}{9} + \frac{{{y^2}}}{4} = 1$

Normal at the point is parallel to the line 

$4x + 2y - 5 = 0$

Slope of normal $=2$

slope of tangent $ = \frac{{ - 1}}{2}$

Point of contact to ellipse $\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} = 1$

and line is $\left( {\frac{{{a^2}m}}{{\sqrt {{a^2}{m^2} + {b^2}} }},\frac{b}{{\sqrt {{a^2}{m^2} + {b^2}} }}} ight)$

Now, ${a^2} = 9,{b^2} = 4$

$	herefore $ point $ = \left( {\frac{{ - 9}}{5},\frac{8}{5}} ight)$

A point on the ellipse, $4x^2 + 9y^2 = 36$, where the normal is parallel to the line, $4x -2y-5 = 0$ , is

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