Suggested languages for you:

Americas

Europe

Q17.

Expert-verifiedFound in: Page 424

Book edition
Middle English Edition

Author(s)
Carter

Pages
804 pages

ISBN
9780079039903

**Identify the coordinates of the vertex and focus, the equations of the axis of symmetry and directrix, and the direction of opening of the parabola with the given equation ${y}^{2}=2x$. Then find the length of latus rectum and graph the parabola.**

The vertex is $\left(0,0\right)$. Focus is $\left(\frac{1}{2},0\right)$. Axis of symmetry is $y=0$. The equation of directrix is $x=-\frac{1}{2}$. The direction of opening of parabola is towards right. The length of latus rectum is $2\text{units}$.

The given equation is ${y}^{2}=2x$.

For two different forms of equations of parabola stated below, use the following key-concept to find vertex, axis of symmetry, focus, directrix, direction of opening of parabola and length of latus rectum.

$\overline{)\begin{array}{ccc}\text{Form of equations}& y=a{\left(x-h\right)}^{2}+k& x=a{\left(y-k\right)}^{2}+h\\ \text{Vertex}& \left(h,k\right)& \left(h,k\right)\\ \text{Axis of symmetry}& x=h& y=k\\ \text{Focus}& \left(h,k+\frac{1}{4a}\right)& \left(h+\frac{1}{4a},k\right)\\ \text{Directrix}& y=k-\frac{1}{4a}& x=h-\frac{1}{4a}\\ \text{Direction of opening}& \begin{array}{l}\text{upward if}a>0,\\ \text{downward if}a<0\end{array}& \begin{array}{l}\text{right if}a>0,\\ \text{left if}a<0\end{array}\\ \text{Length of latus rectum}& \left|\frac{1}{a}\right|\text{units}& \left|\frac{1}{a}\right|\text{units}\end{array}}$

The given equation ${y}^{2}=2x$ is converted to standard form $y=a{\left(x-h\right)}^{2}+k$ as:

$\begin{array}{c}{y}^{2}=2x\mathrm{....}\left(\text{Given}\right)\\ x=\frac{1}{2}{y}^{2}\mathrm{....}\left(\text{Standard form}\right)\end{array}$

Comparing $x=\frac{1}{2}{y}^{2}$ with $x=a{\left(y-k\right)}^{2}+h$, $a=\frac{1}{2},h=0\text{and}k=0$.

The vertex and focus, the equations of the axis of symmetry and directrix, and the direction of opening of the parabola are evaluated as:

The vertex is $\left(0,0\right)$.

Focus is evaluated as:

$\begin{array}{c}\left(h+\frac{1}{4a},k\right)=\left(0+\frac{1}{4\left(\frac{1}{2}\right)},0\right)\\ =\left(\frac{1}{2},0\right)\end{array}$

Axis of symmetry is $y=0$.

The equation for directrix is evaluated as:

$\begin{array}{c}x=h-\frac{1}{4a}\\ x=0-\frac{1}{4\left(\frac{1}{2}\right)}\\ x=-\frac{1}{2}\mathrm{....}\left(\text{Directrix}\right)\end{array}$

The direction of opening of parabola is right because $a>0$.

The length of latus rectum is evaluated as:

$\begin{array}{c}\frac{1}{\left|a\right|}=\frac{1}{\left|\frac{1}{2}\right|}\\ =2\text{units}\end{array}$

The graph of the parabola is shown below.

The vertex is $\left(0,0\right)$. Focus is $\left(\frac{1}{2},0\right)$. Axis of symmetry is $y=0$. The equation of directrix is $x=-\frac{1}{2}$. The direction of opening of parabola is towards right. The length of latus rectum is $2\text{units}$.

94% of StudySmarter users get better grades.

Sign up for free