the normal at any point of hyperbola meets the axes at E,F.find the locus of the midpoint of EF.


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Question Number 46012 by peter frank last updated on 19/Oct/18

$the normal at any point$ $of hyperbola meets the axes$ $at E, F . find the locus$ $of the midpoint of EF .$
	Answered by tanmay.chaudhury50@gmail.com last updated on 20/Oct/18

	$\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1 e q n o f h y p e r b o l a$ $(a s e c θ, b t a n θ) i z a p o i n t o n h y p e r b o l a$ $s l o p e a t (a s e c θ, b t a n θ) . . .$ $\frac{d}{d x} (\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}}) = 0$ $\frac{2 x}{a^{2}} - \frac{2 y}{b^{2}} \times \frac{d y}{d x} = 0$ $\frac{d y}{d x} = \frac{(\frac{- 2 x}{a^{2}})}{(- \frac{2 y}{b^{2}})} = \frac{{x b}^{2}}{{y a}^{2}} s o m = \frac{a s e c θ \times b^{2}}{b t a n θ \times a^{2}} = \frac{b s e c θ}{a t a n θ}$ $s l o p e o f n o r m a l m_{1} s o m \times m_{1} = - 1$ $m_{1} = \frac{- 1}{m} = \frac{- a t a n θ}{b s e c θ}$ $e q n n o r m a l a t (a s e c θ, b t a n θ) i s$ $(y - b t a n θ) = \frac{- a t a n θ}{b s e c θ} (x - a s e c θ)$ $y b s e c θ - b^{2} t a n θ s e c θ = - x a t a n θ + a^{2} t a n θ s e c θ$ $x a t a n θ + y b s e c θ = (a^{2} + b^{2}) t a n θ s e c θ$ $\frac{x}{\frac{(a^{2} + b^{2}) t a n θ s e c θ}{a t a n θ}} + \frac{y}{\frac{(a^{2} + b^{2}) t a n θ s e c θ}{b s e c θ}} = 1$ $s o p o i n t E (\frac{a^{2} + b^{2}}{a} s e c θ, 0) p o i n t F (0, \frac{a^{2} + b^{2}}{b} t a n θ)$ $m i d p o i n t o f E, F (α, β)$ $α = \frac{a^{2} + b^{2}}{2 a} s e c θ β = \frac{a^{2} + b^{2}}{2 b} t a n θ$ $w e k n o w {s e c}^{2} θ - {t a n}^{2} θ = 1$ ${(\frac{2 a α}{a^{2} + b^{2}})}^{2} - {(\frac{2 b β}{a^{2} + b^{2}})}^{2} = 1$ $4 a^{2} α^{2} - 4 b^{2} β^{2} = {(a^{2} + b^{2})}^{2}$ $h e n c e l o c u s i s$ $4 a^{2} x^{2} - 4 b^{2} y^{2} = {(a^{2} + b^{2})}^{2} p l s c h e c k$
	Commented by peter frank last updated on 20/Oct/18

	$yes sir you right thanks$
	Answered by ajfour last updated on 20/Oct/18

	$l e t t h e m i d p o i n t b e (h, k)$ $e q . o f n o r m a l i s :$ $y = - \frac{k x}{h} + 2 k . . . . (i)$ $h y p e r b o l a : \frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1$ $\Rightarrow \frac{d y}{d x} ∣_{x_{1}, y_{1}} = \frac{b^{2}}{a^{2}} (\frac{x_{1}}{y_{1}})$ $e q . o f n o r m a l :$ $y = - \frac{a^{2} y_{1}}{b^{2} x_{1}} (x - x_{1}) + y_{1}$ $\Rightarrow y = - (\frac{a^{2} y_{1}}{b^{2} x_{1}}) x + y_{1} (1 + \frac{a^{2}}{b^{2}}) . . (i i)$ $c o m p a r i n g (i) & (i i)$ $\frac{k}{h} = \frac{a^{2} y_{1}}{b^{2} x_{1}} & 2 k = y_{1} (1 + \frac{a^{2}}{b^{2}})$ $a l s o \frac{x_{1}^{2}}{a^{2}} = 1 + \frac{y_{1}^{2}}{b^{2}}$ $\Rightarrow \frac{k^{2}}{h^{2}} = \frac{a^{4}}{b^{4}} \times \frac{y_{1}^{2}}{a^{2} (1 + \frac{y_{1}^{2}}{b^{2}})}$ $\frac{k^{2}}{h^{2}} = \frac{a^{2}}{b^{2}} \times \frac{1}{(\frac{b^{2}}{y_{1}^{2}} + 1)}$ $\Rightarrow \frac{k^{2}}{h^{2}} = \frac{a^{2}}{b^{2}} \times \frac{1}{[\frac{{(a^{2} + b^{2})}^{2}}{4 b^{2} k^{2}} + 1]}$ $\Rightarrow \frac{k^{2}}{h^{2}} = \frac{a^{2}}{b^{2}} \times \frac{4 b^{2} k^{2}}{[{(a^{2} + b^{2})}^{2} + 4 b^{2} k^{2}]}$ $\Rightarrow 4 b^{2} k^{2} + {(a^{2} + b^{2})}^{2} = 4 a^{2} h^{2}$ $h e n c e r e q u i r e d l o c u s i s$ $4 (a^{2} x^{2} - b^{2} y^{2}) = {(a^{2} + b^{2})}^{2} .$
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