Prove-that-the-perpendicilar-tangent-to-the-ellipse-x-2-a-2-y-2-b-2-1-meets-on-the-circle-x-2-y-2-a-2-b-2-

Question Number 51613 by peter frank last updated on 29/Dec/18

P r o v e t h a t t h e p e r p e n d i c i l a r

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 m e e t s o n t h e

c i r c l e x^{2} + y^{2} = a^{2} + b^{2} .

Answered by peter frank last updated on 29/Dec/18

y = m x + \sqrt{a^{2} m^{2} + b^{2}}

y - m x = \sqrt{a^{2} m^{2} + b^{2}}

{(y - m x)}^{2} = {(\sqrt{a^{2} m^{2} + b^{2}})}^{2}

m^{2} (x^{2} - a^{2}) - 2 m x y + y^{2} - b^{2} = 0

m_{1} + m_{2} = \frac{2 m x}{x^{2} - a^{2}}

m_{1} m_{2} = \frac{y^{2} - b^{2}}{x^{2} - a^{2}}

f o r p e r p e n d i c u l a r

m_{1} m_{2} = - 1

- 1 = \frac{y^{2} - b^{2}}{x^{2} - a^{2}}

y^{2} + x^{2} = a^{2} + b^{2}

Answered by mr W last updated on 29/Dec/18

t w o p e r p e n d i c u l a r l i n e s f r o m p o i n t

(h, k) :

\frac{y - k}{x - h} = m \Rightarrow m x - y - m h + k = 0

\frac{y - k}{x - h} = - \frac{1}{m} \Rightarrow x + m y - h + m k = 0

s i n c e t h e y t a n g e n t t h e e l l i p s e,

a^{2} m^{2} + b^{2} {(- 1)}^{2} - {(- m h + k)}^{2} = 0

\Rightarrow a^{2} m^{2} + b^{2} = m^{2} h^{2} + k^{2} - 2 m h k \dots (i)

a^{2} {(1)}^{2} + b^{2} m^{2} - {(- h + m k)}^{2} = 0

\Rightarrow a^{2} + b^{2} m^{2} = h^{2} + m^{2} k^{2} - 2 m h k \dots (i i)

(i) + (i i) :

\Rightarrow a^{2} (m^{2} + 1) + b^{2} (1 + m^{2}) = h^{2} (m^{2} + 1) + k^{2} (1 + m^{2})

\Rightarrow a^{2} + b^{2} = h^{2} + k^{2}

o r

\Rightarrow x^{2} + y^{2} = a^{2} + b^{2}

Commented by peter frank last updated on 29/Dec/18

t h a n k y o u