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Question Number 51921 by peter frank last updated on 01/Jan/19

A n o r m a l c h o r d t o a n

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

m a k e a n a n g l e o f 45^{°}

w i t h t h e a x i s . p r o v e

t h a t t h e s q u a r e o f i t s

l e n g t h i s e q u a l t o

\frac{32 a^{4} b^{4}}{{(a^{2} + b^{2})}^{3}}

Answered by ajfour last updated on 01/Jan/19

l e t x_{P} = a \cos θ, y_{P} = b \sin θ

- \frac{d x}{d y} ∣_{P} = \frac{a \sin θ}{b \cos θ} = \frac{a}{b} \tan θ = \tan 45 ° = 1

\Rightarrow \tan θ = \frac{b}{a}

\Rightarrow a \cos θ = \frac{a^{2}}{\sqrt{a^{2} + b^{2}}} = \frac{a^{2}}{s} (s a y)

b \sin θ = \frac{b^{2}}{s}

e q . o f n o r m a l

l e t t h i s n o r m a l i n t e r s e c t s e l l i p s e

a o t t h e o t h e r p o i n t Q (h, k)

l e t P Q = l

h = a \cos θ - \frac{l}{\sqrt{2}}, k = b \cos θ - \frac{l}{\sqrt{2}}

o r Q = (\frac{a^{2}}{s} - \frac{l}{\sqrt{2}}, \frac{b^{2}}{s} - \frac{l}{\sqrt{2}})

a s Q i s o n e l l i p s e,

b^{2} {(\frac{a^{2}}{s} - \frac{l}{\sqrt{2}})}^{2} + a^{2} {(\frac{b^{2}}{s} - \frac{l}{\sqrt{2}})}^{2} = a^{2} b^{2}

\Rightarrow b^{2} {(a^{2} \sqrt{2} - s l)}^{2} + a^{2} {(b^{2} \sqrt{2} - s l)}^{2} = 2 a^{2} b^{2} s^{2}

\Rightarrow 2 a^{2} b^{2} (a^{2} + b^{2}) + s^{2} l^{2} (a^{2} + b^{2})

- 2 \sqrt{2} {s a}^{2} b^{2} l - 2 a^{2} b^{2} s^{2} = 0

\Rightarrow s i n c e s^{2} = a^{2} + b^{2}, w e g e t

l^{2} - \frac{2 \sqrt{2} a^{2} b^{2}}{s^{3}} l = 0

\Rightarrow l^{2} = \frac{8 a^{4} b^{4}}{{(a^{2} + b^{2})}^{3}} .

Commented by peter frank last updated on 01/Jan/19

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