Find the equation of the tangents to the ellipse 2x² + y² = 8 which are
i) Parallel to x – 2y – 4 = 0
ii) Perpendicular to x + y + 2 = 0
iii) Which makes an angles $\frac{\mathrm{\pi }}{4}$ with X-axis

Given equation of Ellipse is $\frac{{\mathrm{x}}^2}{4}+\frac{{\mathrm{y}}^2}{8}=1$,
${\mathrm{a}}^2=4,{\mathrm{b}}^2=8$
The equation of any tangent to the Ellipse
$\frac{{\mathrm{x}}^2}{{\mathrm{a}}^2}+\frac{{\mathrm{y}}^2}{{\mathrm{b}}^2}=1\text{ with slope }\mathrm{m}\text{ is }\mathrm{y}=\mathrm{mx}\pm \sqrt{{\mathrm{a}}^2{\mathrm{m}}^2+{\mathrm{b}}^2}$
i) Since tangent is parallel to $\mathrm{x}-2\mathrm{y}=4\Rightarrow \mathrm{m}=\frac{1}{2}$ equation of tangent of $\frac{{\mathrm{x}}^2}{4}+\frac{{\mathrm{y}}^2}{8}=1$ and parallel
to $\mathrm{x}-2\mathrm{y}=4\text{ is }\mathrm{y}=\mathrm{mx}\pm \sqrt{{\mathrm{a}}^2{\mathrm{m}}^2+{\mathrm{b}}^2}$
$\begin{array}{l}\Rightarrow \mathrm{y}=\frac{\mathrm{x}}{2}\pm \sqrt{4\cdot \frac{1}{4}+8}\Rightarrow \mathrm{y}=\frac{\mathrm{x}}{2}\pm 3\Rightarrow 2\mathrm{y}=\mathrm{x}\pm 6\\ \Rightarrow \mathrm{x}-2\mathrm{y}\pm 6=0\end{array}$
ii) Since tangent is perpendicular to x + y + 2 = 0
$\perp$^er slope m = 1
The equation of tangent to ellipse with slope m = 1 is $\mathrm{y}=\mathrm{mx}\pm \sqrt{{\mathrm{a}}^2{\mathrm{m}}^2+{\mathrm{b}}^2}$
$\begin{array}{l}\Rightarrow \mathrm{y}=1.\mathrm{x}\pm \sqrt{4\left(1\right)+8}\Rightarrow \mathrm{y}=\mathrm{x}\pm 2\sqrt{3}\\ \Rightarrow \mathrm{x}-\mathrm{y}\pm 2\sqrt{3}=0\end{array}$
iii) Given slope of tangent  (m) = tan45° = 1
the equation of tangent to ellipse with slope
$\begin{array}{l}\mathrm{m}=1\text{ is }\mathrm{y}=\mathrm{mx}\pm \sqrt{{\mathrm{a}}^2{\mathrm{m}}^2+{\mathrm{b}}^2}\\ \Rightarrow \mathrm{y}=\mathrm{x}\pm \sqrt{4\left(1\right)+8}\Rightarrow \mathrm{y}=\mathrm{x}\pm 2\sqrt{3}\\ \Rightarrow \mathrm{x}-\mathrm{y}\pm 2\sqrt{3}=0\end{array}$