If the eccentricity of the standard hyperbola passing through the point (4,6) is 2, then t…

Question

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:

Accepted Answer

Formula for standard hyperbola :

$${{{x^2}} \over {{a^2}}} - {{{y^2}} \over {{b^2}}} = 1$$

It passes through (4, 6)

$$ \therefore $$ $${{16} \over {{a^2}}} - {{36} \over {{b^2}}} = 1$$ ........(1)

We know, $${e^2} = 1 + {{{b^2}} \over {{a^2}}}$$

$$ \Rightarrow $$ 4 = $$1 + {{{b^2}} \over {{a^2}}}$$ [ as e = 2 ]

$$ \Rightarrow $$ ${{b^2} = 3{a^2}}$ ....(2)

Putting ${{b^2} = 3{a^2}}$ in equation (1)

$${{16} \over {{a^2}}} - {{36} \over {3{a^2}}} = 1$$

$$ \Rightarrow $$ ${{a^2} = 4}$

$$ \therefore $$ , ${{b^2} = 12}$

So Equation of hyperbola is

$${{{x^2}} \over 4} - {{{y^2}} \over {12}} = 1$$

Equation of tangent to the hyperbola at (4, 6) is

$${{4x} \over 4} - {{6y} \over {12}} = 1$$

$$ \Rightarrow $$ $$x - {y \over 2} = 1$$

$$ \Rightarrow $$  2x -- y -- 2 = 0

Solutions · 1