Question

Find the sum of the digits of the product of the perpendiculars from foci on any tangent to the hyperbola $\frac{x^2}{(98{)}^2}-\frac{y^2}{(89{)}^2}=1$.

Answer 1

Given equation of hyperbola is
$\frac{x^2}{(98{)}^2}-\frac{y^2}{(89{)}^2}=1$
Here, $a=98,b=89$
$e=\frac{\sqrt{(98{)}^2+(89{)}^2}}{98}=\frac{\sqrt{17525}}{98}$
So, coordinates of foci are $(\pm \sqrt{17525},0)$
Equation of tangent to the hyperbola is
$y=mx\pm \sqrt{(98{)}^2{m}^2-(89{)}^2}$ ....(1)
Length of perpendicular from $(\sqrt{17525},0)$ on (1) is
$p_1=\left|\frac{m\sqrt{17525}\pm \sqrt{(98{)}^2{m}^2-(89{)}^2}}{\sqrt{1+m^2}}\right|$
Length of perpendicular from $(-\sqrt{17525},0)$ on (1) is
$p_2=\left|\frac{-m\sqrt{17525}\pm \sqrt{(98{)}^2{m}^2-(89{)}^2}}{\sqrt{1+m^2}}\right|$
Now, $p_1{p}_2=\left|\frac{(98{)}^2{m}^2-(89{)}^2-17525m^2}{1+m^2}\right|$
$=\left|\frac{-7921m^2-7921}{1+m^2}\right|$
$=7921$

$\therefore$ the sum of the digits is $7+9+2+1=19$