If the eccentricity of the standard hyperbola | vedclass

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Question

Let equation of hyperbola be $\frac{{{x^2}}}{{{a^2}}} - \frac{{{y^2}}}{{{b^2}}} = 1\,\,$

passes through $\left( {4,6} \right)$

$ \Righ

Vedclass · Accepted Answer

$2x -y -2 = 0$

Vedclass · Answer

Let equation of hyperbola be $\frac{{{x^2}}}{{{a^2}}} - \frac{{{y^2}}}{{{b^2}}} = 1\,\,$

passes through $\left( {4,6} \right)$

$ \Rightarrow \frac{{16}}{{{a^2}}} - \frac{{36}}{{{b^2}}} = 1\,\,\,\,\,\,\,.....\left( i \right)$

Also  ${e^2} = 1 + \frac{{{b^2}}}{{{a^2}}} \Rightarrow {b^2} = 3{a^2}\,\,\,\,\,\,\,\,\,......\left( {ii} \right)$

from $(i)$ and $(ii)$

${a^2} = 4,{b^2} = 12$

equation $\frac{{{x^2}}}{4} - \frac{{{y^2}}}{{12}} = 1\,$

Tangent at $\left( {4,6} \right)$ is $x - \frac{y}{2} = 1\,\,\,$

Or

$2x - y = 2$

If the eccentricity of the standard hyperbola passing, through the point $(4, 6)$ is $2$, then the equation of the tangent to the hyperbola at $(4, 6)$ is

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