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Q55.

Expert-verifiedFound in: Page 789

Book edition
7th Edition

Author(s)
James Stewart, Lothar Redlin, Saleem Watson

Pages
948 pages

ISBN
9781337067508

**Finding the Equation of a Parabola**

**Find an equation of the parabola whose graph is shown**

The equation of parabola is ${y}^{2}=x$

Graph of parabola is given as-

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 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.

Substituting the point $(4,-2)$ in the equation of parabola ${y}^{2}=4px$, we get –

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

$\begin{array}{l}\Rightarrow 4=16p\\ \Rightarrow 16p=4\\ \Rightarrow p=\frac{4}{16}\\ \Rightarrow p=\frac{1}{4}\end{array}$

Now, substituting the value of $p$ in equation ${y}^{2}=4px$ , we get the equation as.

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

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