Question-181394

Question Number 181394 by yaojun2t last updated on 24/Nov/22

Answered by FelipeLz last updated on 25/Nov/22

p : p a r a b o l a

d : d i r e c t r i x

r : l i n e p a s s i n g t h r o u g h F

s : l i n e p a s s i n g t h r o u g h O

{\begin{cases} F = (f, 0) \\ O = (0, 0) \end{cases} \to d : x = - f

p : {(x - f)}^{2} + {(y - 0)}^{2} = {(x - (- f))}^{2} + {(y - y)}^{2}

p : {(x - f)}^{2} + y^{2} = {(x + f)}^{2}

p : y^{2} = 4 f x

A \in p \to A = (a, 2 \sqrt{a f})

B \in p \to B = (b, - 2 \sqrt{b f})

r : y = m x + n

F \in r \to 0 = m f + n \Rightarrow n = - m f

A \in r \to 2 \sqrt{a f} = m a - m f \Rightarrow m = \frac{2 \sqrt{a f}}{a - f}

B \in r \to - 2 \sqrt{b f} = m b - m f \Rightarrow m = \frac{2 \sqrt{b f}}{f - b}

m = m

\frac{2 \sqrt{a f}}{a - f} = \frac{2 \sqrt{b f}}{f - b}

f \sqrt{a} - b \sqrt{a} = a \sqrt{b} - f \sqrt{b}

f (\sqrt{a} + \sqrt{b}) = a \sqrt{b} + b \sqrt{a}

f (\sqrt{a} + \sqrt{b}) = \sqrt{a b} (\sqrt{a} + \sqrt{b})

f = \sqrt{a b}

s : y = u x + v

O \in s \to 0 = u \times 0 + v \Rightarrow v = 0

A \in s \to 2 \sqrt{a f} = u a + 0 \Rightarrow u = 2 \sqrt{\frac{f}{a}} = 2 \sqrt[4]{\frac{b}{a}}

D = (- f, 2 \sqrt[4]{\frac{b}{a}} \times (- f))

D = (- \sqrt{a b}, - 2 \sqrt[4]{\frac{b}{a}} \times \sqrt{a b})

D = (- \sqrt{a b}, - 2 \sqrt[4]{\frac{b}{a}} \times \sqrt[4]{a^{2} b^{2}})

D = (- \sqrt{a b}, - 2 \sqrt[4]{{a b}^{3}})

B = (b, - 2 \sqrt{b f}) = (b, - 2 \sqrt[4]{{a b}^{3}})

y_{B} = y_{D}

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