Find the axis, vertex, focus, equation of directrix, latus rectum and length of the latus rectum for the parabola $y^2 + 8x - 6y + 1 = 0$. Also sketch its graph.
Given equation of parabola is: $\;\;$ $y^2 + 8x - 6y + 1 = 0$
i.e. $y^2 - 6y = - 8x - 1$
i.e. $y^2 - 6y + 9 = - 8x - 1 + 9$
i.e. $\left(y - 3\right)^2 = - 8 x + 8$
i.e. $\left(y - 3\right)^2 = - 8 \left(x - 1\right)$ $\;\;\; \cdots \; (1)$
Let $X = x - 1$ $\;$ and $\;$ $Y = y - 3$
Then equation $(1)$ can be written as
$Y^2 = - 8 X$
i.e. $\left(Y - 0\right)^2 = - 4 \times 2 \times \left(X - 0\right)$ $\;\;\; \cdots \; (2)$
Comparing equation $(2)$ with the standard equation $\;\;$ $\left(Y - k\right)^2 = - 4 \times a \times \left(X - h\right)$ $\;$ gives
$\left(h, k\right) = \left(0, 0\right)$ $\;$ and $\;$ $a = 2$
$\therefore$ $\;$ For the given parabola,
| Referred to X, Y | Referred to x, y $x = X + 1$; $\;$ $y = Y + 3$ |
|
|---|---|---|
| Axis of symmetry | X axis $\;$ $\left(Y = 0\right)$ | $Y = 0 \implies y = 3$ i.e. $\;$ $y - 3 = 0$ |
| Vertex | $V = \left(h, k\right) = \left(0,0\right)$ | $X = 0 \implies x = 1$ $Y = 0 \implies y = 3$ $\therefore$ $\;$ $V = \left(1,3\right)$ |
| Focus | $F = \left(-a, 0\right) = \left(-2, 0\right)$ | $X = -2 \implies x = -1$ $Y = 0 \implies y = 3$ $\therefore$ $\;$ $F = \left(-1, 3\right)$ |
| Equation of directrix | $X = a$ i.e. $\;$ $X = 2$ |
$X = 2 \implies x = 3$ i.e. $\;$ $x - 3 = 0$ |
| Equation of latus rectum | $X = -a$ i.e. $\;$ $X = -2$ |
$X = -2 \implies x = -1$ i.e. $\;$ $x + 1 = 0$ |
| Length of latus rectum | $4a = 4 \times 2 = 8$ | $8$ |
