Chapter 10.2, Problem 57AYU

In Problems 57-64, write an equation for each parabola.

To determine

The equation for the given parabola.

Answer to Problem 57AYU

${\left(y-1\right)}^2=-\left(x-2\right)$

Explanation of Solution

Given:

Formula used:

Vertex	Focus	Directrix	Equation	Description
$\left(h, k\right)$	$\left(h-a, k\right)$	$x=h+a$	$\left(y-k\right)²=-4a\left(x-h\right)$	Parabola, axis of symmetry is parallel to $x\text{-axis}$, opens left.

Calculation:

The given graph opens left, the vertex $V\left(h, k\right)$ is given by $V\left(2, 1\right)$ and it passes through \left(1,0\right). The equation for the parabola which opens left and with vertex $\left(h, k\right)$ is given by $\left(y-k\right)²=-4a\left(x-h\right)$.

Substituting $\left(2, 1\right)$ in the above equation we get,

$\left(y-1\right)²=-4a\left(x-2\right)$

Parabola passes through $\left(1, 0\right)$ substituting $\left(1, 0\right)$ in the equation above, we get,

${\left(0-1\right)}^2=-4a*\left(1-2\right)$

$1=-4a\left(-1\right)$

$a=\frac{1}{4}$, substituting $a=\frac{1}{4}$ in $\left(y-1\right)²=-4a\left(x-2\right)$.

{\left(y-1\right)}^2=-4*\frac{1}{4}*\left(x-2\right)

${\left(y-1\right)}^2=-\left(x-2\right)$

The equation of the parabola is given by, ${\left(y-1\right)}^2=-\left(x-2\right)$.