Find the equation of tangent to the ellipse x^2 +4y^2 =4 which are perpendicular to the line 2x−3y=1 ∗merry X−mas and happy new year∗


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Question Number 51151 by peter frank last updated on 24/Dec/18

$F i n d t h e e q u a t i o n o f$ $t a n g e n t t o t h e e l l i p s e$ $x^{2} + 4 y^{2} = 4 w h i c h a r e$ $p e r p e n d i c u l a r t o t h e$ $l i n e 2 x - 3 y = 1$ $* m e r r y X - m a s a n d h a p p y n e w y e a r *$
	Answered by tanmay.chaudhury50@gmail.com last updated on 24/Dec/18

	$3 y = 2 x - 1$ $y = \frac{2}{3} x - \frac{1}{3}$ $t a n g e n t y = - \frac{3}{2} x + c$ $c = \sqrt{a^{2} m^{2} + b^{2}}$ $\frac{x^{2}}{2^{2}} + \frac{y^{2}}{1^{2}} = 1$ $c = \sqrt{2^{2} \times \frac{{(- 3)}^{2}}{2^{2}} + 1^{2}} = \sqrt{10}$ $t a n g e n t i s$ $y = - \frac{3 x}{2} + \sqrt{10}$
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