The normal at the point P(4cos θ,3sin θ) on the ellipse (x^2 /(16)) +(y^2 /9)=1 meets the x−axis and y−axis at A and B respectively show that locus of the mid−point of AB is an ellipse with the same eccentricity as given ellipse.


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Question Number 62753 by peter frank last updated on 24/Jun/19

$T h e n o r m a l a t t h e p o i n t$ $P (4 \cos θ, 3 \sin θ) o n t h e$ $e l l i p s e \frac{x^{2}}{16} + \frac{y^{2}}{9} = 1 m e e t s$ $t h e x - a x i s a n d y - a x i s$ $a t A a n d B r e s p e c t i v e l y$ $s h o w t h a t l o c u s o f t h e$ $m i d - p o i n t o f A B i s a n$ $e l l i p s e w i t h t h e s a m e$ $e c c e n t r i c i t y a s g i v e n$ $e l l i p s e .$
	Answered by Hope last updated on 25/Jun/19

	$e q n n o r m a l (y - 3 s i n θ) = {(\frac{- d x}{d y})}_{(4 c o s θ, 3 s i n θ)} (x - 4 c o s θ)$ $\frac{x^{2}}{16} + \frac{y^{2}}{9} = 1$ $\frac{2 x \times \frac{d x}{d y}}{16} + \frac{2 y}{9} = 0$ $\frac{x \times \frac{d x}{d y}}{8} = \frac{- 2 y}{9} \to \frac{d x}{d y} = \frac{- 16 y}{9 x} = \frac{- 16 \times 3 s i n θ}{9 \times 4 c o s θ} = \frac{- 4}{3} t a n θ$ $(y - 3 s i n θ) = \frac{4 t a n θ}{3} (x - 4 c o s θ)$ $p u t x = 0$ $y - 3 s i n θ = \frac{4 s i n θ}{3 c o s θ} \times - 4 c o s θ$ $y = 3 s i n θ - \frac{16 s i n θ}{3} = \frac{- 7 s i n θ}{3}$ $B (0, \frac{- 7 s i n θ}{3})$ $p u t y = 0$ $- 3 s i n θ = \frac{4 s i n θ}{3 c o s θ} (x - 4 c o s θ)$ $x = \frac{- 9 s i n θ c o s θ}{4 s i n θ} + 4 c o s θ$ $x = \frac{- 9 c o s θ + 16 c o s θ}{4} = \frac{7 c o s θ}{4} A (\frac{7 c o s θ}{4}, 0)$ $M i d p o i n t o f A B = (\frac{7 c o s θ}{8}, \frac{- 7 s i n θ}{6})$ $l o c u s α = \frac{7 c o s θ}{8} β = \frac{- 7 s i n θ}{6}$ ${(\frac{8 α}{7})}^{2} + {(\frac{6 β}{- 7})}^{2} = 1$ $l o c u s \frac{x^{2}}{{(\frac{7}{8})}^{2}} + \frac{y^{2}}{{(\frac{7}{6})}^{2}} = 1$ $e c c e n r i x i t y = \frac{\sqrt{{(\frac{7}{6})}^{2} - {(\frac{7}{8})}^{2}}}{\frac{7}{6}} = \frac{6}{7} \times 7 \times \sqrt{\frac{64 - 36}{8^{2} \times 6^{2}}}$ $e_{2} = \frac{6}{7} \times 7 \times \frac{1}{8 \times 6} \times 2 \sqrt{7} = \frac{\sqrt{7}}{4}$ $\frac{x^{2}}{16} + \frac{y^{2}}{9} = 1 e_{1} = \frac{\sqrt{16 - 9}}{4} = \frac{\sqrt{7}}{4}$ $s o e_{1} = e_{2} = \frac{\sqrt{7}}{4} p r o v e d$
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