Question

If the value's of $m$ for which the line $y=mx+2\sqrt{5}$ touches the hyperbola $16x^2-9y^2=144$ are roots of the equation $x^2-(a+b)x-4=0$, then the value of $a+b=$

Answer 1

$\because y=mx+2\sqrt{5}$ is tangent to hyperbola
$\frac{x^2}{9}-\frac{y^2}{16}=1$
$\therefore (2\sqrt{5}{)}^2=a^2{m}^2-b^2$
$\Rightarrow 9m^2-16=20$
$9m^2=36\Rightarrow m_1,m_2=\pm 2$
As we know that $m_1,m_2$ are the roots of
$x^2-(a+b)x-4=0\therefore m_1+m_2=a+b\Rightarrow a+b=0$