The-tangent-at-P-to-an-ellipse-meets-directrix-at-Q-prove-that-the-line-joining-the-corresponding-focus-to-P-and-Q-are-perpendicular-

Question Number 51588 by peter frank last updated on 28/Dec/18

T h e t a n g e n t a t P t o a n e l l i p s e

m e e t s d i r e c t r i x a t Q

p r o v e t h a t t h e l i n e

j o i n i n g t h e c o r r e s p o n d i n g

f o c u s t o P a n d Q a r e

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

Answered by peter frank last updated on 28/Dec/18

f r o m t a n g e n t

a y s i n θ + b x \cos θ - a b = 0 \dots (i)

P = (a \cos θ, b \sin θ)

Q = (\frac{a}{e}, y) \Rightarrow d i r e c t r i x

S = (a e, 0)

a y s i n θ + b x \cos θ - a b = 0

p u t x = \frac{a}{e}

a y s i n θ + b \times \frac{a}{e} \cos θ - a b = 0

y = \frac{- b (\cos θ - e)}{e \sin θ}

Q = (\frac{a}{e}, \frac{- b (\cos θ - e)}{e \sin θ})

S = (a e, 0)

s l o p e \Rightarrow m_{S Q} = \frac{- b (\cos θ - e)}{a \sin θ (1 - e^{2})}

P = (a \cos θ, b \sin θ)

S = (a e, 0)

s l o p s \Rightarrow m_{S P} = \frac{b \sin θ}{a \cos θ - a e}

S Q a n d S P a r e p e r p e n d i c u l a

{s l o p e}_{S Q} \times {s l o p e}_{S P} = - 1

m_{S Q} = [\frac{- b (\cos θ - e)}{a \sin θ (1 - e^{2})}] \times [\frac{b \sin θ}{a \cos θ - a e}]

= - \frac{b^{2}}{a^{2} (1 - e^{2})} [b^{2} = a^{2} (1 - e^{2})

= - \frac{b^{2}}{b^{2}} = - 1

h e n c e s h o w n