Question

A parabola has the origin as its focus and the line x = 2 as the directrix. Then the vertex of the parabola is at: -
(a) (0, 1)
(b) (2, 0)
(c) (0, 2)
(d) (1, 0)

Answer 1

Hint: Assume the equation of the parabola as ${{y}^{2}}=4a\left( x-h \right)$. Substitute (x – h) = X and write the equation as ${{y}^{2}}=4aX$. Now, compare this equation of parabola with the most common form ${{y}^{2}}=4ax$ whose directrix is x = -a, focus is S(a, 0) and the vertex is (0, 0)that is the origin. For the shifted parabola ${{y}^{2}}=4a\left( x-h \right)$, substitute (x – h) in place of x for the relations of directrix and focus. Find the values of ‘a’ and ‘h’ and substitute in the coordinates of vertex to get the answer.

 

Complete step by step solution:

Here we have been provided with the equation of directrix and coordinate of the focus of the parabola and we are asked to find the coordinate of its vertex.

Now, we can see that the directrix is x = 2 that means the parabola is of the form ${{y}^{2}}=4ax$ and its vertex must lie on the x axis. Also we have the focus as (0, 0) that means there is a shift in the parabola either left or right along the x axis. We have concluded this because we know that the focus of the parabola ${{y}^{2}}=4ax$ lies at (a, 0) and its vertex lies at the origin. Since there is shift in x axis so let us consider the equation of the required parabola as ${{y}^{2}}=4a\left( x-h \right)$.

Considering (x – h) = X we get the equation as:

$\Rightarrow {{y}^{2}}=4aX$

So, comparing the equation ${{y}^{2}}=4aX$ with ${{y}^{2}}=4ax$ we have (x – h) in place of x. We know that directrix of ${{y}^{2}}=4ax$ is x = -a so for the parabola ${{y}^{2}}=4aX$ the directrix will be given as:

$ \Rightarrow X=-a $

$ \Rightarrow x-h=-a $

$ \Rightarrow x=h-a $

In the question we have been provided with the directrix as x = 2, so substituting it in the above relation we get,

$\Rightarrow 2=h-a.............\left( i \right)$

Now, the focus of ${{y}^{2}}=4ax$ lies at (a, 0) that means at x = a so the focus of ${{y}^{2}}=4aX$ will lie at:

$ \Rightarrow X=a $

$  \Rightarrow x-h=a $

$ \Rightarrow x=h+a $

In the question we have been given that focus is at origin that means at x = 0, so substituting it in the above relation we get,

$\Rightarrow 0=h+a.............\left( ii \right)$

Solving equations (i) and (ii) we get,

$\Rightarrow a=-1$ and $h=1$

At last, we know that the vertex of ${{y}^{2}}=4ax$ lies at (0, 0) that means at x = 0 so the vertex of ${{y}^{2}}=4aX$ must be at:

$ \Rightarrow X=0 $

$\Rightarrow x-h=0 $

$ \Rightarrow x=h $

Substituting the obtained value of h obtained above we get,

$\therefore x=1$

Therefore, the coordinates of the vertex of the given parabola is (1, 0). Hence option (d) is the correct answer.

 

Note: Here we haven’t found anything about the y – coordinate because we already know that it will be 0 for the values of focus and vertex. If the parabola would have been of the type ${{x}^{2}}=4ay$ then we would have performed every function on the y coordinate as in that case x would have been 0 for the values of focus and directrix. As we can see that the value of ‘a’ is negative and ‘h’ is positive that means the required parabola opens leftward and the shift is towards right.