Find the equation of tangent to the curve $x^2-2y^2=8$ which are perpendicular to the line $x-y+29=0$
Slope of the line $x-y+29=0$ is $m=1$
Since the required tangent is perpendicular to the given line
slope of tangent $=-\dfrac{1}{m}=-1$ $\;\; \cdots$ (1)
Differentiating the equation of the curve $x^2-2y^2=8$ with respect to x gives
$2x-4y \dfrac{dy}{dx}=0$
i.e. $2y \dfrac{dy}{dx}=x$
$\implies$ $\dfrac{dy}{dx}=\dfrac{x}{2y}$ $\;\; \cdots$ (2)
$\therefore$ $\;$ We have from equations (1) and (2)
$\dfrac{x}{2y}=-1$
$\implies$ $x=-2y$ $\;\; \cdots$ (3)
In view of equation (3), the equation of the curve becomes
$\left(-2y\right)^2-2y^2=8$
i.e. $2y^2=8$ $\implies$ $y^2=4$ $\implies$ $y=\pm 2$
Substituting the value of y in equation (3) gives
$x=-2y=\mp 4$
$\therefore$ $\;$ The tangents to the given curve are located at the points $\left(-4,2\right)$, $\left(4,-2\right)$
$\therefore$ $\;$ Equation of tangent at $\left(-4,2\right)$ is
$y-2=-1\left(x+4\right)$ $\implies$ $x+y+2=0$
$\therefore$ $\;$ Equation of tangent at $\left(4,-2\right)$ is
$y+2=-1\left(x-4\right)$ $\implies$ $x+y-2=0$
$\therefore$ $\;$ The two tangents to the given curve are $x+y+2=0$ and $x+y-2=0$