a-Normal-to-any-point-on-the-hyperbola-XY-C-meet-the-x-axis-at-A-and-tangents-meets-the-y-axis-at-B-find-the-locus-of-the-mid-point-of-AB-b-find-the-equation-of-assymptotes-of-i-x-2-4-y-2-5-1

Question Number 50979 by peter frank last updated on 23/Dec/18

a) N o r m a l t o a n y p o i n t o n

t h e h y p e r b o l a X Y = C

m e e t t h e x - a x i s a t A

a n d t a n g e n t s m e e t s

t h e y - a x i s a t B . f i n d t h e

l o c u s o f t h e m i d p o i n t o f A B

b) f i n d t h e e q u a t i o n o f

a s s y m p t o t e s o f

(i) \frac{x^{2}}{4} - \frac{y^{2}}{5} = 1

(i i) \frac{{(x - 1)}^{2}}{16} - \frac{{(y - 3)}^{2}}{9} = 1

Commented by ggururajguru0219@gmail.com last updated on 23/Dec/18

v g 5 n

Answered by mr W last updated on 23/Dec/18

(a)

x y = c

y^{'} = - \frac{c}{x^{2}}

P (h, \frac{c}{h})

y^{'} = - \frac{c}{h^{2}}

t a n g e n t :

y = \frac{c}{h} - \frac{c}{h^{2}} (x - h)

B (0, y_{B})

\Rightarrow y_{B} = \frac{2 c}{h}

n o r m a l :

y = \frac{c}{h} + \frac{h^{2}}{c} (x - h)

A (x_{A}, 0)

0 = \frac{c}{h} + \frac{h^{2}}{c} (x_{A} - h)

\Rightarrow x_{A} = h - \frac{c^{2}}{h^{3}}

m i d p o i n t o f A B : (p, q)

q = \frac{y_{B}}{2} = \frac{1}{2} \times \frac{2 c}{h} = \frac{c}{h}

\Rightarrow h = \frac{c}{q}

p = \frac{x_{A}}{2} = \frac{1}{2} (h - \frac{c^{2}}{h^{3}}) = \frac{1}{2} (\frac{c}{q} - \frac{q^{3}}{c})

2 c p q = c^{2} - q^{4}

\Rightarrow 2 c x y = c^{2} - y^{4}

Commented by mr W last updated on 23/Dec/18

Commented by peter frank last updated on 23/Dec/18

t h a n k y o u

Commented by peter frank last updated on 23/Dec/18

s i r p l e a s e c h e c k t h i r d l i n e

P (h, \frac{c}{h}) i t s h o u l d b e P (c h, \frac{c}{h})

Commented by mr W last updated on 23/Dec/18

w h a t i s y o u r c u r v e ?

x y = c o r x y = c^{2} ?

i f x y = c a s t h e q u e s t i o n s a y s, t h e n

t h e p o i n t i s P (h, \frac{c}{h}) b e c a u s e h \times \frac{c}{h} = c .

Commented by peter frank last updated on 23/Dec/18

v e r y s o r r y s i r f o u n d m y m i s t a k e . m y

x y = c^{2} i n s t e a d y o f x y = c

Answered by peter frank last updated on 23/Dec/18

b) \frac{x^{2}}{4} - \frac{y^{2}}{5} = 1

a = 2 b = \sqrt{5}

y = \pm \frac{b}{a} x

y = \pm \frac{\sqrt{5}}{2}

i i) \frac{{(x - 1)}^{2}}{16} - \frac{{(y - 3)}^{2}}{9} = 1

a = 4 b = 3 p = 1 q = 3

y - q = \pm \frac{b}{a} (x - p)

y - 3 = \pm \frac{3}{4} (x - 1)

y = \frac{3}{4} x - \frac{9}{4} o r - \frac{3}{4} x + \frac{9}{4}

Answered by peter frank last updated on 23/Dec/18

a) e q u a t i o n o f n o r m a l

a t x y = c

y - \frac{{x t}^{2}}{c} + \frac{t^{3}}{c} - \frac{c}{t} = 0

A (x, 0) \Rightarrow A (t - \frac{c^{2}}{t^{3}}, 0)

f r o m t a n g e n t a t x y = c

{y t}^{2} + c x - 2 c t = 0

B (0, y) \Rightarrow (0, \frac{2 c}{t})

A (t - \frac{c^{2}}{t^{3}}, 0), B (0, \frac{2 c}{t})

x, y) = (\frac{x_{1} + x_{2}}{2}, \frac{y_{1} + y_{2}}{2})

(x, y) = (\frac{c t}{2} - \frac{c}{2 t^{3}}), \frac{c}{y}

x = \frac{c}{2} - \frac{c}{2 t^{3}} \dots \dots (i)

t = \frac{c}{y} \dots \dots . (i i)

s u b i i i n i

2 x = \frac{c}{y} - \frac{y^{3}}{c}

y^{4} + 2 x c y - c^{2} = 0

Commented by mr W last updated on 23/Dec/18

p l e a s e c h e c k e q n . o f n o r m a l :

y - {x t}^{2} + {c t}^{3} - \frac{c}{t} = 0

i t s h o u l d b e, i t h i n k,

y - \frac{{x t}^{2}}{c} + \frac{t^{3}}{c} - \frac{c}{t} = 0

Commented by peter frank last updated on 23/Dec/18

M r w p l e a s e c h e c k r i g h t o r w r o n g

Commented by mr W last updated on 23/Dec/18

p l e a s e c h e c k e q n . o f t a n g e n t :

{y t}^{2} + x - 2 c t = 0 p p

i t s h o u l d b e, i t h i n k,

{y t}^{2} + c x - 2 c t = 0

Commented by mr W last updated on 23/Dec/18

p l e a s e c h e c k :

t h e f i n a l e q n . s h o u l d b e, i t h i n k,

y^{4} + 2 c x y - c^{2} = 0

Commented by peter frank last updated on 23/Dec/18

e q u a t i o n o f t a n g e n t a n d n o r m q l

t o x y = c a t (c t, \frac{c}{t})

x = c t

y = \frac{c}{t}

y^{^{'}} = - \frac{1}{t^{2}}

\frac{y - \frac{c}{t}}{x - c t} = - \frac{1}{t^{2}}

{y t}^{2} + x - 2 c t = 0

n o r m a l t o x y = c

y^{^{'}} = t^{2}

\frac{y - \frac{c}{t}}{x - c t} = t^{2}

{x t}^{2} - {c t}^{3} = y - \frac{c}{t}

{x t}^{2} - {c t}^{3} + \frac{c}{t} = y

Commented by peter frank last updated on 23/Dec/18

M r w s i r p l e a s e c h e c k

Commented by mr W last updated on 23/Dec/18

i f x y = c, t h e n

a t p o i n t (t, \frac{c}{t}), n o t a t (c t, \frac{c}{t})

y^{'} = - \frac{c}{x^{2}} = - \frac{c}{t^{2}}

\dots \dots

Commented by peter frank last updated on 23/Dec/18

y o u r a b s o l u t e r i g h t