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Question Number 51489 by peter frank last updated on 27/Dec/18

G i v e n t h a t y = m x + c

i s 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 \frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1

f i n d c o o r d i n a t e o f

p o i n t o f c o n t a c t .

Answered by ajfour last updated on 27/Dec/18

P (a \cos θ, b \sin θ)

\frac{x \cos θ}{a} + \frac{y \sin θ}{b} - 1 = 0 (t a n g e n t)

b x \cos θ + a y \sin θ - a b = 0

o r m x - y + c = 0

\Rightarrow \frac{b \cos θ}{m} = \frac{a \sin θ}{- 1} = \frac{- a b}{c}

\tan θ = - \frac{b}{a m}

\frac{b^{2}}{c^{2}} + \frac{a^{2} m^{2}}{c^{2}} = \sin^{2} θ + \cos^{2} θ = 1

\Rightarrow \sqrt{a^{2} m^{2} + b^{2}} = ∣ c ∣

a \cos θ = a (\frac{- a m}{\sqrt{a^{2} m^{2} + b^{2}}})

b \sin θ = b (\frac{b}{\sqrt{a^{2} m^{2} + b^{2}}})

P (- \frac{a^{2} m}{c}, \frac{b^{2}}{c}) .

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