એક અતિવલયની નાભિઓ ( pm 2,0) અને તેની ઉત્કેન્દ | vedclass

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Question

$\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$

$ae =2$ and $e =\frac{3}{2} \Rightarrow a =\frac{4}{3}$

also $b ^2= a ^2 e ^2- a ^2 \Rightarrow 4-\frac{16}{

Vedclass · Accepted Answer

$12$

Vedclass · Answer

$\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$

$ae =2$ and $e =\frac{3}{2} \Rightarrow a =\frac{4}{3}$

also $b ^2= a ^2 e ^2- a ^2 \Rightarrow 4-\frac{16}{9}$

$\Rightarrow b^2=\frac{20}{9}$

$	ext { Slope of tangent }=\frac{3}{2}$

So tangent equation will be

$y = mx \pm \sqrt{a^2 m^2-b^2}$

$\Rightarrow y =\frac{3 x}{2} \pm \sqrt{\frac{16}{9} \cdot \frac{9}{4}-\frac{20}{9}}$

$\Rightarrow y =\frac{3 x}{2} \pm \frac{4}{3} \Rightarrow\left| x _{	ext {intercept }}ight|=\frac{8}{9}$

$\left|y_{	ext {intercept }}ight|=\frac{4}{3}$

$\Rightarrow|6 a|+|5 b|=\frac{48}{9}+\frac{60}{9}=\frac{109}{9}=12$

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અતિવલય $ \frac{{{x^2}}}{{16}}\,\, - \,\,\frac{{{{\left( {y\,\, - \,\,2} \right)}^2}}}{9}\,\, = \,\,1\,$ ની નાભીઓ.......

આપેલ અતિવલય માટે નાભિઓ, શિરોબિંદુઓ, ઉત્કેન્દ્રતા અને નાભિલંબની લંબાઈ મેળવો: $\frac{x^{2}}{9}-\frac{y^{2}}{16}=1$