A normal to the hyperbola, 4x^2 - 9y^2, = 36 | vedclass

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Question

Given, $4{x^2} - 9{y^2} = 36$

ater differentiating w.r.t.$x$, we get 

$4.2x - 9.2y.\frac{{dy}}{{dx}} = 0$

$ \Rightarrow $ slope of

Vedclass · Accepted Answer

$9x^2 -4y^2\, = 169$

Vedclass · Answer

Given, $4{x^2} - 9{y^2} = 36$

ater differentiating w.r.t.$x$, we get 

$4.2x - 9.2y.\frac{{dy}}{{dx}} = 0$

$ \Rightarrow $ slope of tangent $ = \frac{{dy}}{{dx}} = \frac{{4x}}{{9y}}$

so, slope of normal $ = \frac{{ - 9y}}{{4x}}$

Now, equation of normal at point $({x_0},{y_0})$ is given by

${y_0} - {y_0} = \frac{{ - 9y}}{{4x}}\left( {{x_0} - {x_0}} ight)$

As normal intersects $X$ axis at $A$, Then

$A \equiv \left( {\frac{{13{x_0}}}{9},0} ight)$ 

and $B \equiv \left( {0,\frac{{13{y_0}}}{4}} ight)$

As $OABP$ is a parallelogram 

$	herefore $ midpoint of $OB \equiv \left( {0,\frac{{13{y_0}}}{8}} ight) \equiv $ Midpoint of $AP$

So,$P\left( {x,y} ight) \equiv \left( {\frac{{ - 13{x_0}}}{9},\frac{{13{y_0}}}{4}} ight)\,\,\,\,\,\,......\left( i ight)$

$	herefore \,({x_0},{y_0})$ lies on hyperbola, therfore 

$4{\left( {{x_0}} ight)^2} - 9{\left( {{y_0}} ight)^2} = 36\,\,\,\,\,\,\,\,.......\left( {ii} ight)$

Feom equation $(i):$ ${x_0} = \frac{{ - 9x}}{{13}}\,$ and ${y_0} = \frac{{4y}}{{13}}$

From equation $(ii)$, we get 

$9{x^2} - 4{y^2} = 169$

Hence, locus point $P$ is :$9{x^2} - 4{y^2} = 169$

A normal to the hyperbola, $4x^2 - 9y^2\, = 36$ meets the co-ordinate axes $x$ and $y$ at $A$ and $B$, respectively . If the parallelogram $OABP$ ( $O$ being the origin) is formed, then the locus of $P$ is

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