The value of m, for which the line y = mx + f | vedclass

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Question

(b) We know that the equation of the normal of the conic ,

$\frac{{{x^2}}}{{{a^2}}} - \frac{{{y^2}}}{{{b^2}}} = 1$ at point $(a\sec 	heta ,\,b	an

Vedclass · Accepted Answer

$ - \frac{2}{{\sqrt 3 }}$

Vedclass · Answer

(b) We know that the equation of the normal of the conic ,

$\frac{{{x^2}}}{{{a^2}}} - \frac{{{y^2}}}{{{b^2}}} = 1$ at point $(a\sec 	heta ,\,b	an 	heta )$

is $ax\sec 	heta + by\cot 	heta = {a^2} + {b^2}$

or $y = \frac{{ - a}}{b}\sin 	heta \,x + \frac{{{a^2} + {b^2}}}{{b\cot 	heta }}$

Comparing above equation with equation $y = mx + \frac{{25\sqrt 3 }}{3}$ and taking $a = 4,\,b = 3$

we get, $\frac{{{a^2} + {b^2}}}{{b\cot 	heta }} = \frac{{25\sqrt 3 }}{3}$

$ \Rightarrow $ $	an 	heta = \sqrt 3$

$\Rightarrow 	heta = {60^o}$

and $m = - \frac{a}{b}\sin 	heta = \frac{{ - 4}}{3}\sin {60^o}$

= $\frac{{ - 4}}{3} 	imes \frac{{\sqrt 3 }}{2} = \frac{{ - 2}}{{\sqrt 3 }}$.

The value of $m$, for which the line $y = mx + \frac{{25\sqrt 3 }}{3}$, is a normal to the conic $\frac{{{x^2}}}{{16}} - \frac{{{y^2}}}{9} = 1$, is

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