Find the equations of tangent and normal to the ellipse $x^2 + 4y^2 = 32$ at $\theta = \dfrac{\pi}{4}$
Equation of ellipse is $\;$ $x^2 + 4y^2 = 32$
i.e. $\;$ $\dfrac{x^2}{32} + \dfrac{y^2}{32 / 4} = 1$
i.e. $\;$ $\dfrac{x^2}{32} + \dfrac{y^2}{8} = 1$ $\;\;\; \cdots \; (1)$
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
$a^2 = 32 \implies a = 4 \sqrt{2}$ $\;$ and $\;$ $b^2 = 8 \implies b = 2 \sqrt{2}$
$\theta = \dfrac{\pi}{4}$ represents the point $\left(a \cos \theta, b \sin \theta\right)$
i.e. the point $\left(4 \sqrt{2} \cos \left(\dfrac{\pi}{4}\right), 2 \sqrt{2} \sin \left(\dfrac{\pi}{4}\right)\right)$ $\;$ i.e. $\left(4,2\right)$
Equation of tangent to the ellipse at the point $\left(x_1, y_1\right)$ is $\;$ $\dfrac{xx_1}{a^2} + \dfrac{yy_1}{b^2} = 1$
Equation of normal to the ellipse at the point $\left(x_1, y_1\right)$ is $\;$ $\dfrac{a^2 x}{x_1} - \dfrac{b^2 y}{y_1} = a^2 - b^2$
Here $\left(x_1, y_1\right) = \left(4,2\right)$
$\therefore$ $\;$ The required equation of tangent to the ellipse is
$\dfrac{4x}{32} + \dfrac{2y}{8} = 1$
i.e. $\;$ $\dfrac{x}{8} + \dfrac{y}{4} = 1$
i.e. $\;$ $x + 2y - 8 = 0$
The required equation of normal to the ellipse is
$\dfrac{32x}{4} - \dfrac{8y}{2} = 32 - 8$
i.e. $\;$ $8x - 4y = 24$
i.e. $\;$ $2x - y - 6 = 0$
ALTERNATELY
Equation of tangent to the ellipse $\dfrac{x^2}{a^2} + \dfrac{y^2}{b^2} = 1$ in parametric form is $\;$ $\dfrac{x \cos \theta}{a} + \dfrac{y \sin \theta}{b} = 1$
Equation of normal to the ellipse $\dfrac{x^2}{a^2} + \dfrac{y^2}{b^2} = 1$ in parametric form is $\;$ $\dfrac{ax}{\cos \theta} - \dfrac{b y}{\sin \theta} = a^2 - b^2$
$\therefore$ $\;$ Required equation of tangent is
$\dfrac{x \cos \left(\dfrac{\pi}{4}\right)}{4 \sqrt{2}} + \dfrac{y \sin \left(\dfrac{\pi}{4}\right)}{2 \sqrt{2}} = 1$
i.e. $\;$ $\dfrac{x}{8} + \dfrac{y}{4} = 1$
i.e. $\;$ $x + 2y - 8 = 0$
Required equation of normal is
$\dfrac{4 \sqrt{2} x}{\cos \left(\dfrac{\pi}{4}\right)} - \dfrac{2 \sqrt{2} y}{\sin \left(\dfrac{\pi}{4}\right)} = 32 - 8$
i.e. $\;$ $8x - 4y = 24$
i.e. $\;$ $2x - y - 6 = 0$