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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} \boldsymbol A (\frac{7 c o s θ}{4}, 0)

M i d p o i n t o f A \boldsymbol 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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