Prove-that-line-lx-my-n-0-is-tangent-to-the-ellipse-x-2-a-2-y-2-b-2-1-if-a-2-l-2-b-2-m-2-n-2-

Question Number 77127 by peter frank last updated on 03/Jan/20

P r o v e t h a t l i n e l x + m y + n = 0

i s 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 i f a^{2} l^{2} + b^{2} m^{2} = n^{2}

Answered by jagoll last updated on 03/Jan/20

s u p p o s e a > b \Leftrightarrow t a n g e n t l i n e

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

e q u a l t o l x + m y + n = 0 \Rightarrow

y = - \frac{l}{m} x - \frac{n}{m}

m = - \frac{l}{m} \Rightarrow m^{2} = - l

- \frac{n}{m} = - \sqrt{a^{2} m^{2} + b^{2}}

\Rightarrow \frac{n^{2}}{m^{2}} = a^{2} m^{2} + b^{2}

n^{2} = (- l) (- {l a}^{2} + b^{2})

n^{2} = l^{2} a^{2} - {l b}^{2} \Rightarrow n^{2} = l^{2} a^{2} + m^{2} b^{2}

Commented by peter frank last updated on 03/Jan/20

t h a n k y o u

Answered by mr W last updated on 03/Jan/20

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

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

m^{2} b^{2} x^{2} + a^{2} {(l x + n)}^{2} = m^{2} a^{2} b^{2}

m^{2} b^{2} x^{2} + a^{2} (l^{2} x^{2} + n^{2} + 2 l n x) = m^{2} a^{2} b^{2}

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

d u e t o t a n g e n c y :

Δ = {(2 a^{2} l n)}^{2} - 4 (a^{2} l^{2} + b^{2} m^{2}) a^{2} (n^{2} - b^{2} m^{2}) = 0

a^{2} l^{2} n^{2} - (a^{2} l^{2} + b^{2} m^{2}) (n^{2} - b^{2} m^{2}) = 0

(- n^{2} + a^{2} l^{2} + b^{2} m^{2}) b^{2} m^{2} = 0

\Rightarrow a^{2} l^{2} + b^{2} m^{2} = n^{2}

Commented by peter frank last updated on 03/Jan/20

t h a n k y o u