Let P(asec theta ,;btan theta ) and Q(asec va | vedclass

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Question

(d) Given $P$ $(a\sec 	heta ,\,b	an 	heta )$ and $Q\,(a\sec \phi ,\,b	an \phi )$

The equation of tangent at point $P$ is $\frac{{x\sec 	heta }

Vedclass · Accepted Answer

$ - \left( {\frac{{{a^2} + {b^2}}}{b}} \right)$

Vedclass · Answer

(d) Given $P$ $(a\sec 	heta ,\,b	an 	heta )$ and $Q\,(a\sec \phi ,\,b	an \phi )$

The equation of tangent at point $P$ is $\frac{{x\sec 	heta }}{a} - \frac{{y	an 	heta }}{b} = 1$

$m$ of tangent $ = \frac{b}{{	an 	heta }} 	imes \frac{{\sec 	heta }}{a} = \frac{b}{a}.\frac{1}{{\sin 	heta }}$

Hence the equation of perpendicular at $P$ is

$y - b	an 	heta = - \frac{{a\sin 	heta }}{b}(x - a\sec 	heta )$

or $by - {b^2}	an 	heta = - a\sin 	heta \,x + {a^2}	an 	heta $

or $a\sin 	heta \,x + by = ({a^2} + {b^2})	an 	heta $.....$(i)$

Similarly the equation of perpendicular at $Q$ is

$a\sin \phi \,x + by = ({a^2} + {b^2})	an \phi $&hellip;..$(ii)$

On multiplying $(i)$ by $\sin \phi $ and $(ii)$ by $\sin 	heta $

$a\sin 	heta \sin \phi \,x + b\sin \phi y = ({a^2} + {b^2})	an 	heta \sin \phi $

$a\sin \phi \sin 	heta \,x + b\sin 	heta y = ({a^2} + {b^2})	an \phi \sin 	heta $

On subtraction by,

$(\sin \phi - \sin 	heta ) = ({a^2} + {b^2})(	an 	heta \sin \phi - 	an \phi \sin 	heta )$

$	herefore y = k = \frac{{{a^2} + {b^2}}}{b}.\frac{{	an 	heta \sin \phi - 	an \phi \sin 	heta }}{{\sin \phi - \sin 	heta }}$

$\because 	heta  + \phi  = \frac{\pi }{2}$ ==> $\phi = \frac{\pi }{2} - 	heta $

==> $\sin \phi = \cos 	heta $ and $	an \phi = \cot 	heta $

$	herefore y = k = \frac{{{a^2} + {b^2}}}{b}.\frac{{	an 	heta \cos 	heta - \cot 	heta \sin 	heta }}{{\cos 	heta - \sin 	heta }}$

$ = \frac{{{a^2} + {b^2}}}{b}\left( {\frac{{\sin 	heta - \cos 	heta }}{{\cos 	heta - \sin 	heta }}} ight) = - \frac{{({a^2} + {b^2})}}{b}$.

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