Question

How do you find the vertex, the coordinates of focus, and the directrix of the parabola $10x = {y^2}$ ?

Answer 1

From the question, we can say that it is a concave right parabola. This is so because the question is in the form of ${y^2} = 4ax$. Now, we can find out the vertex, focus, and the directrix of the parabola according to the form given.

Complete step by step solution:
As we know that the parabola is a concave right parabola because of the form it is having. The form is:
${y^2} = 4ax$
We will find the vertex first. We can see that here any transformations are not taking place in the parabola. Therefore, the vertex is $(0,0)$. Now, we will find the focus of the parabola. The form of the focus of parabola is $(a,0)$. From the formula ${y^2} = 4ax$, we know that the $a$ here is the focus of the parabola. So, we need to find the $a$ from the question given. According to the formula we get that:
$10x = 4ax$

We can rewrite this equation as:
$\Rightarrow 4ax = 10x$
Now, we have to shift $4x$to the other side, so that $a$ is alone, and we can get the focus of the parabola.
$\Rightarrow a = \dfrac{{10x}}{{4x}}$
Now, the terms which are divisible or are similar gets cancelled here, and then we get:
$\Rightarrow a = \dfrac{5}{2}$
Hence, according to the frame of the focus of the parabola, we get that the coordinates of the focus of the parabola is $\left( {\dfrac{5}{2},0} \right)$.
Now, we are going to discover the directrix of the parabola. The equation to discover the directrix of the parabola is:
$x = - a$
We know that $a = \dfrac{5}{2}$. Therefore, $x = - \dfrac{5}{2}$.

So, the directrix of the parabola is $- \dfrac{5}{2}$.

Note: In this question there were no transformations in the parabola so the answer came as $(0,0)$. Otherwise, the formula for vertex is $y = a{(x - h)^2} + k$ where $(h,k)$ are known as the vertex. This was the formula to find the vertex of Quadratic Equations.