Find the equations of tangent and normal to the parabola $x^2 + 2x - 4y + 4 = 0$ at $\left(0,1\right)$.
Equation of given parabola is: $\;$ $x^2 + 2x - 4y + 4 = 0$
i.e. $\;$ $x^2 + 2x = 4y - 4$
i.e. $\;$ $x^2 + 2x +1 = 4y - 4 + 1$
i.e. $\;$ $\left(x + 1\right)^2 = 4y - 3$
i.e. $\;$ $\left(x + 1\right)^2 = 4 \left(y - \dfrac{3}{4}\right)$ $\;\;\; \cdots \; (1)$
Let $\;$ $X = x + 1$ $\;\;\; \cdots \; (2a)$ $\;$ and $\;$ $Y = y - \dfrac{3}{4}$ $\;\;\; \cdots \; (2b)$
Then, we have from equations $(1)$, $(2a)$ and $(2b)$,
$X^2 = 4Y$ $\;\;\; \cdots \; (3)$
Comparing equation $(3)$ with the standard equation of parabola $\;$ $X^2 = 4a Y$ $\;$ gives $\;$ $a = 1$
Given: Point $\left(x_1, y_1\right) = \left(0, 1\right)$
$\therefore$ $\;$ We have from equations $(2a)$ and $(2b)$,
when $x_1 = 0 \implies X_1 = 0 + 1 = 1$
and when $y_1 = 1 \implies Y_1 = 1 - \dfrac{3}{4} = \dfrac{1}{4}$
$\therefore$ $\;$ $\left(X_1, Y_1\right) = \left(1, \dfrac{1}{4}\right)$
Equation of tangent at $\left(X_1, Y_1\right)$ to the parabola given by equation $(3)$ is
$XX_1 = 2a \left(Y + Y_1\right)$
$\therefore$ $\;$ Equation of tangent to the parabola given by equation $(3)$ at the point $\left(X_1, Y_1\right)$ is
$1 \times X = 2 \times 1 \times \left(Y + \dfrac{1}{4}\right)$
i.e. $\;$ $X = 2Y + \dfrac{1}{2}$ $\;\;\; \cdots \; (4)$
Equation of normal at $\left(X_1, Y_1\right)$ to the parabola given by equation $(3)$ is
$YX_1 + 2a X = X_1 Y_1 + 2a X_1$
i.e. $\;$ $1 \times Y + 2 \times 1 \times X = 1 \times \dfrac{1}{4} + 2 \times 1 \times 1$
i.e. $\;$ $Y + 2X = \dfrac{9}{4}$ $\;\;\; \cdots \; (5)$
Substituting for $X$ and $Y$ from equations $(2a)$ and $(2b)$ in equation $(4)$, the equation of tangent to parabola given by equation $(1)$ is
$x +1 = 2 \left(y - \dfrac{3}{4}\right) + \dfrac{1}{2}$
i.e. $\;$ $x = 2y + \dfrac{1}{2} - \dfrac{3}{2} - 1$
i.e. $\;$ $x - 2y + 2 = 0$
Substituting for $X$ and $Y$ from equations $(2a)$ and $(2b)$ in equation $(5)$, the equation of normal to the parabola given by equation $(1)$ is
$y - \dfrac{3}{4} + 2 \times \left(x + 1\right) = \dfrac{9}{4}$
i.e. $\;$ $2x + y - \dfrac{3}{4} + 2 - \dfrac{9}{4} = 0$
i.e. $\;$ $2x + y - 1 = 0$