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Q22.

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 $y={x}^{2}-12x+20$. Then find the length of latus rectum and graph the parabola.**

The vertex is $\left(6,-16\right)$. Focus is $\left(6,-\frac{63}{4}\right)$. Axis of symmetry is $x=6$. The equation of directrix is $y=-\frac{65}{4}$. The direction of opening of parabola is upwards. The length of latus rectum is $1\text{unit}$.

The given equation is $y={x}^{2}-12x+20$.

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 $y={x}^{2}-12x+20$ is converted to standard form $y=a{\left(x-h\right)}^{2}+k$ as:

$\begin{array}{c}y={x}^{2}-12x+\mathrm{20....}\left(\text{Given}\right)\\ y={x}^{2}-12x+20+{\left(-6\right)}^{2}-{\left(-6\right)}^{2}\mathrm{....}\left(\text{Add and subtract}{\left(-6\right)}^{2}\right)\\ y=\left({x}^{2}-12x+{\left(-6\right)}^{2}\right)+\left(20-{\left(-6\right)}^{2}\right)\\ y={\left(x-6\right)}^{2}-\mathrm{16....}\left(\text{Standard form}\right)\end{array}$

Comparing $y={\left(x-6\right)}^{2}-16$ with $y=a{\left(x-h\right)}^{2}+k$, $a=1,h=6\text{and}k=-16$.

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 $\left(6,-16\right).....\left(\because h=6,k=-16\right)$

Focus is evaluated as:

$\begin{array}{c}\left(h,k+\frac{1}{4a}\right)=\left(6,-16+\frac{1}{4\left(1\right)}\right)\\ =\left(6,-\frac{63}{4}\right)\end{array}$

Axis of symmetry is$x=6....\left(\because h=6\right)$.

The equation for directrix is evaluated as:

$\begin{array}{c}y=k-\frac{1}{4a}\\ y=-16-\frac{1}{4\left(1\right)}\\ y=-\frac{65}{4}\mathrm{....}\left(\text{Directrix}\right)\end{array}$

The direction of opening of parabola is upwards because \[a>0\].

The length of latus rectum is evaluated as:

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

The graph of the parabola is shown below.

The vertex is $\left(6,-16\right)$. Focus is $\left(6,-\frac{63}{4}\right)$. Axis of symmetry is $x=6$. The equation of directrix is $y=-\frac{65}{4}$. The direction of opening of parabola is upwards. The length of latus rectum is $1\text{unit}$.

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