Question

Finding the Equation of a Parabola

Find an equation of the parabola whose graph is shown

Short answer

The equation of parabola is $\Rightarrow x^2=2(\sqrt{5}-1)y$

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 –

$x^2=4py$

Where, $p>0$ then parabola open upwards

$p<0$ then parabola open downwards

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 $x^2=4py$ to obtain an equation of a parabola.

Step 3. Calculation

The coordinate of focus is $(0,p)$ and $p>0$.

The point of intersection of line and parabola whose abscissa is 2 and the ordinate is calculated by putting abscissa value 2 in the equation of a parabola $x^2=4py$.

${\left(2\right)}^2=4py$

$\begin{array}{l}\Rightarrow 4py=4\\ \Rightarrow y=\frac{1}{p}\end{array}$

So, the point of intersection of line and parabola is $\left(2,\frac{1}{p}\right)$.

The slope for the line is $\frac{1}{2}$ .

On applying the slope formula for two points $(0,p)$ and $\left(2,\frac{1}{p}\right)$ , we get –

$\begin{array}{l}\frac{1}{2}=\frac{\frac{1}{p}-p}{2-0}\\ \Rightarrow \frac{\frac{1-p^2}{p}}{2}=\frac{1}{2}\\ \Rightarrow \frac{1-p^2}{p}=1\\ \Rightarrow 1-p^2=p\\ \Rightarrow 1-p^2-p=0\\ \Rightarrow -p^2-p+1=0\\ \Rightarrow p^2+p-1=0\end{array}$

Applying the quadratic formula for the equation $p^2+p-1=0$-

$\begin{array}{l}p=\frac{-1\pm \sqrt{{\left(1\right)}^2-4\left(1\right)(-1)}}{2\left(1\right)}\\ p=\frac{-1\pm \sqrt{1+4}}{2}\\ p=\frac{-1\pm \sqrt{5}}{2}\end{array}$

As the parabola opens upwards, so –

$p=\frac{-1+\sqrt{5}}{2}$

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

$\begin{array}{l}{x}^2=4\left(\frac{\sqrt{5}-1}{2}\right)y\\ \Rightarrow x^2=2(\sqrt{5}-1)y\end{array}$