ઉત્કેન્દ્ર્તા mathrm{e} વાળા એક અતિવલયનાં નાભ | vedclass

Name: PDF Generator App | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

Given $\frac{2 \mathrm{~b}^2}{\mathrm{a}}=9$ and $\frac{\mathrm{a}}{\mathrm{e}}= \pm \frac{4}{\sqrt{3}}$

equation of tangent $y-\sqrt{3} x+\sqrt{3}

Vedclass · Accepted Answer

$61$

Vedclass · Answer

Given $\frac{2 \mathrm{~b}^2}{\mathrm{a}}=9$ and $\frac{\mathrm{a}}{\mathrm{e}}= \pm \frac{4}{\sqrt{3}}$

equation of tangent $y-\sqrt{3} x+\sqrt{3}=0$

by equation of tangent

Let slope $=\mathrm{S}=\sqrt{3}$

Constant $=-\sqrt{3}$

By condition of tangency

$ \Rightarrow 6=6 \mathrm{a}^2-9 \mathrm{a} $

$ \Rightarrow \mathrm{a}=2, \mathrm{~b}^2=9$

Equation of Hyperbola is

$\frac{x^2}{4}-\frac{y^2}{9}=1$ and for tangent

Point of contact is $(4,3 \sqrt{3})=\left(\mathrm{x}_0, \mathrm{y}_0ight)$

Now $\mathrm{e}=\sqrt{1+\frac{9}{4}}=\frac{\sqrt{13}}{2}$

Again product of focal distances

$\mathrm{m}=\left(\mathrm{x}_0 \mathrm{e}ight. $$ +\mathrm{a})\left(\mathrm{x}_0 \mathrm{e}-\mathrm{a}ight) $

$\mathrm{m}+4 \mathrm{e}^2 $$ =20 \mathrm{e}^2-\mathrm{a}^2 $

$ =20 	imes \frac{13}{4}-4=61$

(There is a printing mistake in the equation of directrix $x= \pm \frac{4}{\sqrt{3}}$.

Corrected equation is $\mathrm{x}= \pm \frac{4}{\sqrt{13}}$ for directrix, as eccentricity must be greater than one, so question must be bonus)

Similar Questions

રેખા ${\text{2x}}\,\, + \;\,\sqrt {\text{6}} y\,\, = \,\,2$ એ વક્ર $\,{x^2}\, - \,\,2{y^2}\,\, = \,\,4\,\,$ ને કયા બિંદુ આગળ સ્પર્શે છે?

$\frac{{{x^2}}}{{{a^2}}}\,\, - \,\,\frac{{{y^2}}}{{{b^2}}}\,\, = \,\,1\,$ ના અનંતસ્પર્શકો વચ્ચેનો ખૂણો ${\text{ = }}\,...........$