If-the-tangents-at-the-end-of-a-focal-chord-of-parabola-meet-the-tangent-at-the-vertex-in-C-D-prove-that-CD-substends-a-right-angle-at-the-focus-

Question Number 118482 by peter frank last updated on 17/Oct/20

If the tangents at the

end of a focal chord of

parabola meet the

tangent at the vertex

in C, D . prove that CD

substends a right angle

at the focus

Answered by mr W last updated on 18/Oct/20

s a y v e r t e x i s t h e o r i g i n, t h e f a c u s i s

F (0, h), t h e n t h e e q n . o f p a r a b o l a i s

y = \frac{x^{2}}{4 h}

s a y t h e f o c a l c h o r d i s

y = h + m x

i n t e r s e c t i o n w i t h p a r a b o l a :

\frac{x^{2}}{4 h} = h + m x

x^{2} - 4 m h x - 4 h^{2} = 0

\Rightarrow x = 2 h (m \pm \sqrt{m^{2} + 1})

e n d p o i n t A :

x_{A} = 2 h (m - \sqrt{m^{2} + 1})

y_{A} = h + 2 h m (m - \sqrt{m^{2} + 1})

t a n g e n t a t A :

\frac{d y}{d x} = \frac{x}{2 h} = m - \sqrt{m^{2} + 1}

y = h + 2 h m (m - \sqrt{m^{2} + 1}) + (m - \sqrt{m^{2} + 1}) [x - 2 h (m - \sqrt{m^{2} + 1})]

0 = h + 2 h m (m - \sqrt{m^{2} + 1}) + (m - \sqrt{m^{2} + 1}) [x - 2 h (m - \sqrt{m^{2} + 1})]

\Rightarrow x = x_{C} = - \frac{h}{m - \sqrt{m^{2} + 1}} - 2 h \sqrt{m^{2} + 1}

e n d p o i n t B :

x_{B} = 2 h (m + \sqrt{m^{2} + 1})

y_{B} = h + 2 h m (m + \sqrt{m^{2} + 1})

t a n g e n t a t B :

y = h + 2 h m (m + \sqrt{m^{2} + 1}) + (m + \sqrt{m^{2} + 1}) [x - 2 h (m + \sqrt{m^{2} + 1})]

0 = h + 2 h m (m + \sqrt{m^{2} + 1}) + (m + \sqrt{m^{2} + 1}) [x - 2 h (m + \sqrt{m^{2} + 1})]

\Rightarrow x = x_{D} = - \frac{h}{m + \sqrt{m^{2} + 1}} + 2 h \sqrt{m^{2} + 1}

i n c l i n a t i o n o f C F :

m_{C F} = \frac{h}{0 - x_{C}} = \frac{1}{\frac{1}{m - \sqrt{m^{2} + 1}} + 2 \sqrt{m^{2} + 1}}

= \frac{m - \sqrt{m^{2} + 1}}{1 + 2 \sqrt{m^{2} + 1} (m - \sqrt{m^{2} + 1})}

= \frac{m - \sqrt{m^{2} + 1}}{2 m \sqrt{m^{2} + 1} - 2 m^{2} - 1}

i n c l i n a t i o n o f D F :

m_{D F} = \frac{h}{0 - x_{D}} = \frac{1}{\frac{1}{m + \sqrt{m^{2} + 1}} - 2 \sqrt{m^{2} + 1}}

= \frac{m + \sqrt{m^{2} + 1}}{1 - 2 \sqrt{m^{2} + 1} (m + \sqrt{m^{2} + 1})}

= - \frac{m + \sqrt{m^{2} + 1}}{2 m \sqrt{m^{2} + 1} + 2 m^{2} + 1}

m_{C F} \times m_{D F} = - \frac{m - \sqrt{m^{2} + 1}}{2 m \sqrt{m^{2} + 1} - 2 m^{2} - 1} \times \frac{m + \sqrt{m^{2} + 1}}{2 m \sqrt{m^{2} + 1} + 2 m^{2} + 1}

= - \frac{m^{2} - (m^{2} + 1)}{{(2 m \sqrt{m^{2} + 1})}^{2} - {(2 m^{2} + 1)}^{2}}

= - \frac{- 1}{4 m^{4} + 4 m^{2} - 4 m^{4} - 4 m^{2} - 1}

= - 1

\Rightarrow C F ⊥ D F

Commented by peter frank last updated on 18/Oct/20

diagram please

Commented by peter frank last updated on 18/Oct/20

thank you . i appriaciate

your time .

Commented by mr W last updated on 18/Oct/20

Commented by peter frank last updated on 18/Oct/20

GODBLESS YOU

Commented by peter frank last updated on 18/Oct/20

where this came from

y = h + 2 h m (m - \sqrt{m^{2} + 1}) + (m - \sqrt{m^{2} + 1}) [x - 2 h (m - \sqrt{m^{2} + 1})]

Commented by mr W last updated on 18/Oct/20

p o i n t C (x_{C}, y_{C})

x_{C} = 2 h (m - \sqrt{m^{2} + 1})

y_{C} = h + {m x}_{C} = h + 2 h m (m - \sqrt{m^{2} + 1})

t h e i n c l i n a t i o n o f t a n g e n t a t C i s

m_{C} = m - \sqrt{m^{2} + 1}

\Rightarrow t h e e q n . o f t a n g e n t a t C i s

y = y_{C} + m_{C} (x - x_{C})

\Rightarrow y = h + 2 h m (m - \sqrt{m^{2} + 1}) + (m - \sqrt{m^{2} + 1}) [x - 2 h (m - \sqrt{m^{2} + 1})]

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