Question

Check whether the following equation represent parabola and if they do so, then find their vertex, focus, directrix, axis and latus rectum:${y^2} - 8y - x + 19 = 0$

Answer 1

Here we will convert the given equation into a standard form of parabola. The standard form of parabola is given as: ${(y - k)^2} = 4a(x - h)$. After converting, we will compare with the standard form and find the vertex (h, k), length of latus rectum (4a) and equation of directrix for the given equation.

Complete step-by-step answer:
The given Equation is: ${y^2} - 8y - x + 19 = 0$.
We can rewrite it as, ${y^2} - 8y = x - 19$.
On adding and subtracting 16 on left hand side, we get:
${y^2} - 8y + 16 - 16 = x - 19$
We know that $({a^2} + {b^2} - 2ab) = {(a - b)^2}$. So the above expression can be written as:
${(y - 4)^2} - 16 = x - 19$
$\Rightarrow {(y - 4)^2} = x - 3$ (1)
We know that the general form of a parabola is given as: ${(y - k)^2} = 4a(x - h)$, where (h, k) is the vertex and 4a is the length of the latus rectum.
And the equation of directrix is given as: x + a –h =0.
Therefore, on comparing the equation 1 with the general equation of parabola, we get:
Vertex (h, k) = (3, 4)
4a = 1 = length of latus rectum.
And the equation of directrix is
x + a –h =0.
On putting the value of a and h, we have:
$x + \dfrac{1}{4} - 3 = 0$
$\Rightarrow \dfrac{{4x + 1 - 12}}{4} = 0$
$\Rightarrow$ 4x – 11 =0
Therefore, the vertex for the given parabola is (3, 4) and the length of the latus rectum is 1 and directrix is 4x – 11 =0.

Note: In these types of questions always try to convert the equation in a known form. Using that form we can find the vertex, focus and directrix. You should also know that focus is at (h + a, k) for the parabola ${(y - k)^2} = 4a(x - h)$ and the ends of latus rectum is $\left( {h + a,k \pm 2a} \right)$.