Question

Finding the Equation of a Parabola

Find an equation of the parabola whose graph is shown

Short answer

The equation of parabola is $x^2=-4\sqrt{2}y$

Step-by-step solution

Step 1. Given information

Graph of parabola is given as-

Step 2. Concept used

The equation of a parabola having a vertex at origin is –

$y^2=4px$

Where, $p>0$ then parabola opens to the right

$p<0$ then parabola opens to the left.

The length of a rectangle is equal to focal diameter and the breadth of a rectangle is equal to focal length.

The directrix is the right side of $y$-axis is so, which means the focus should be negative, the parabola opens to the left and the coordinates of focus are obtained. On substituting the value of \[p\] in the equation $y^2=4px$ to obtain an equation of a parabola.

Step 3. Calculation

The focus coordinate is $(0,p),p<0$.

Let the length and breadth of the rectangle be $l$ and $b$ respectively.

$l=4p$

The breadth of a rectangle is equal to focal length, so –

$b=p$

The area of a rectangle$=8$.

On applying the formula for the area of the square, we get –

$\begin{array}{l}l\times b=8\\ \Rightarrow 4p\times p=8\\ \Rightarrow 4p^2=8\\ \Rightarrow p^2=2\\ \Rightarrow p=\pm \sqrt{2}\end{array}$

As the parabola opens downwards, so –

$p=-\sqrt{2}$

The coordinate for the point of intersection of directrix and $x$-axis is$(-4,0)$.

The equation of directrix is-

$x=-4$

The coordinate of focus will be $(4,0)$, so $p=4$.

Now, substituting the value of $p$ in equation $x^2=4py$ , we get the equation as.

$\begin{array}{l}{x}^2=4(-\sqrt{2})y\\ \Rightarrow x^2=-4\sqrt{2}y\end{array}$