माना अतिपरवलय frac{x^{2}}{9}-frac{y^{2}}{4}=1 | vedclass

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Question

Let the coordinate at point of intersection of  normal at $P$ and $Q$ be $(h,k)$ 

Since, equation of normal to the hyperbola $\,\fra

Vedclass · Accepted Answer

$-\frac{13}{2}$

Vedclass · Answer

Let the coordinate at point of intersection of  normal at $P$ and $Q$ be $(h,k)$ 

Since, equation of normal to the hyperbola $\,\frac{{{x^2}}}{{{a^2}}} - \frac{{{y^2}}}{{{b^2}}} = 1$ At point $\left( {{x_1},{y_1}} ight)$ is 

$\frac{{{a^2}x}}{{{x_1}}} + \frac{{{b^2}y}}{{{y_1}}} = {a^2} + {b^2}$ therefore equation of normal to the hyperbola $\frac{{{x^2}}}{{{3^2}}} - \frac{{{y^2}}}{{{2^2}}} = 1$

at point $P\left( {3\,\sec 	heta ,2\,	an 	heta } ight)$ is

$\frac{{{3^2}x}}{{3\,\sec 	heta }} + \frac{{{2^2}y}}{{2	an 	heta }} = {3^2} + {2^2}$

$ \Rightarrow \boxed{3x\,\cos 	heta  + 2y\cot 	heta  = {3^2} + {2^2}} ......\left( 1 ight)$

Similarly , equation of normal to the hyperbola $\frac{{{x^2}}}{{{3^2}}} - \frac{{{y^2}}}{{{2^2}}}$ at point 

$Q\left( {3\,\sec \phi ,2\,	an \phi } ight)$ is

$\frac{{{3^2}x}}{{3\,\sec \phi }} + \frac{{{2^2}y}}{{2	an \phi }} = {3^2} + {2^2}$

$ \Rightarrow \boxed{3x\,\cos \phi  + 2y\cot \phi  = {3^2} + {2^2}} ......\left( 2 ight)$

Given $	heta  + \phi  = \frac{\pi }{2} \Rightarrow \phi  = \frac{\pi }{2} - 	heta $ and these passes through $(h,k)$

$	herefore $ Frem eq. $(2)$

$ \Rightarrow \boxed{3h\,\sin 	heta  + 2k	an 	heta  = {3^2} + {2^2}} ......\left( 3 ight)$ 

and $\boxed{3h\,\cos 	heta  + 2k\cot 	heta  = {3^2} + {2^2}} ......\left( 4 ight)$

Commparing equation $(3) \;and \;(4)$, we get

$3h\,\cos 	heta  + 2k\cot 	heta  = 3h\,\sin 	heta  + 2k	an 	heta $

$3h\,\cos 	heta  - 3h\sin 	heta  = 2k	an 	heta  - 2k\cot 	heta $

$3h\left( {\cos 	heta  - \sin 	heta } ight) = 2k\left( {	an 	heta  - \cot 	heta } ight)$

$3h\left( {\cos 	heta  - \sin 	heta } ight)$

$ = 2k\frac{{\left( {\sin 	heta  - \cos 	heta } ight)\left( {\sin 	heta  + \cos 	heta } ight)}}{{\sin 	heta \cos 	heta }}$

or $3h = \frac{{ - 2k\left( {\sin 	heta \cos 	heta } ight)}}{{\sin 	heta \cos 	heta }} ......\left( 5 ight)$

Now, putting the value of equation $(5)$ in eq.$(3)$

$\frac{{ - 2k\left( {\sin 	heta \cos 	heta } ight)\sin 	heta }}{{\sin 	heta \cos 	heta }} + \,2k	an 	heta  = {3^2} + {2^2}$

$ \Rightarrow \,2k	an 	heta  - 2k + 2k	an 	heta  = 13$

$ - 2k = 13 \Rightarrow k = \frac{{ - 13}}{2}$

Hence, ordinate of point of intersection  of the at $P$ and $Q$ is $\frac{{ - 13}}{2}$

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