If a hyperbola passes through the point P(10, 16) and it has vertices at (± 6, 0), then the equation of the normal to it at P is :

Let hyperbola is ${{{x^2}} \over {{a^2}}} - {{{y^2}} \over {{b^2}}} = 1$ <br><br>vertices ($\pm$a, 0 ) = ($\pm$6, 0) $\Rightarrow$ a = 6 <br><br>Hyperbola passes through p(10, 16) <br><br>$\therefore$ ${{{{10}^2}} \over {{6^2}}} - {{{{16}^2}} \over {{b^2}}} = 1$ <br><br>$\Rightarrow$ b = 12 <br><br>$\therefore$ Required hyperbola is ${{{x^2}} \over {36}} - {{{y^2}} \over {144}} = 1$ <br><br>Equation of normal will be <br><br>${{{a^2}x} \over {{x_1}}} + {{{b^2}y} \over {{y_1}}} = {a^2} + {b^2}$ <br><br>At P(10,16) normal is <br><br>${{36x} \over {10}} + {{144y} \over {16}} = 36 + 144$ <br><br>$\Rightarrow$ 18x + 45y = 900 <br><br>$\Rightarrow$ 2x + 5y = 100