Find the equation of tangents to the ellipse $\dfrac{x^2}{20}+ \dfrac{y^2}{5} = 1$, which are perpendicular to $x+y+2=0$.
Equation of ellipse is $\;$ $\dfrac{x^2}{20} + \dfrac{y^2}{5} = 1$ $\;\;\; \cdots \; (1)$
The major axis of the ellipse given by equation $(1)$ is along the X axis.
Comparing equation $(1)$ with the standard equation of ellipse $\;$ $\dfrac{x^2}{a^2} + \dfrac{y^2}{b^2} = 1$ $\;$ gives
$a^2 = 20$ $\;$ and $\;$ $b^2 = 5$
Equation of given line is $\;$ $x + y + 2 = 0$
i.e. $\;$ $y = -x - 2$
Slope of given line $= m_1 = -1$
$\because$ $\;$ the required tangents are perpendicular to the given line,
slope of tangent $= m = \dfrac{-1}{m_1} = 1$
Equations of tangents (with slope $m$) to the ellipse are $\;$ $y = mx \pm \sqrt{a^2 m^2 + b^2}$
$\therefore$ $\;$ Required equations of tangents to the ellipse are
$y = 1 \times x \pm \sqrt{20 \times 1^2 + 5}$
i.e. $\;$ $y = x \pm 5$