If a hyperbola passes through the point P(10,16) and it has vertices at (6,0), then the eq…

Question

If a hyperbola passes through the point $P(10,16)$ and it has vertices at $(\pm6,0)$, then the equation of the normal to it at $P$ is:

Accepted Answer

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

vertices ($$ \pm $$a, 0 ) = ($$ \pm $$6, 0) $$ \Rightarrow $$ a = 6

Hyperbola passes through p(10, 16)

$$ \therefore $$ $${{{{10}^2}} \over {{6^2}}} - {{{{16}^2}} \over {{b^2}}} = 1$$

$$ \Rightarrow $$ b = 12

$$ \therefore $$ Required hyperbola is $${{{x^2}} \over {36}} - {{{y^2}} \over {144}} = 1$$

Equation of normal will be

$${{{a^2}x} \over {{x_1}}} + {{{b^2}y} \over {{y_1}}} = {a^2} + {b^2}$$

At P(10,16) normal is

$${{36x} \over {10}} + {{144y} \over {16}} = 36 + 144$$

$$ \Rightarrow $$ 18x + 45y = 900

$$ \Rightarrow $$ 2x + 5y = 100

Solutions · 1