# How do you find the standard form of the equation for the parabola with vertex (4,0) and passing through the point (1,2)?

May 14, 2018

Equation is $y = \frac{2}{9} {x}^{2} - \frac{16}{9} x + \frac{32}{9}$ or $x = - \frac{3}{4} {y}^{2} + 4$

#### Explanation:

With vertex at $\left(h , k\right)$, we can have two types of parabolas - (i) of regular vertical type $y = a {\left(x - h\right)}^{2} + k$ and (ii) of horizontal type $x = a {\left(y - k\right)}^{2} + h$. In standard form it is $y = a {x}^{2} + b x + c$ or $x = a {y}^{2} + b y + c$

If it is regular type, if vertex is $\left(4 , 0\right)$, our equation is

$y = a {\left(x - 4\right)}^{2} + 0$ and as it passes through $\left(1 , 2\right)$, we have

$2 = a {\left(1 - 4\right)}^{2}$ or $9 a = 2$ i.e. $a = \frac{2}{9}$ and

equation is $y = \frac{2}{9} {\left(x - 4\right)}^{2}$ or $9 y = 2 {\left(x - 4\right)}^{2}$

or in standard form equation is $y = \frac{2}{9} {x}^{2} - \frac{16}{9} x + \frac{32}{9}$

If parabola is horizontal, if vertex is $\left(4 , 0\right)$, our equation is

$x = a {\left(y - 0\right)}^{2} + 4$ and as it passes through $\left(1 , 2\right)$, we have

$1 = a \cdot {2}^{2} + 4$ or $4 a = - 3$ i.e. $a = - \frac{3}{4}$ and

equation is $x = - \frac{3}{4} {y}^{2} + 4$ or $4 x + 3 {y}^{2} = 16$

and in standard form equation is $x = - \frac{3}{4} {y}^{2} + 4$
graph{(4x+3y^2-16)(2(x-4)^2-9y)=0 [-10, 10, -5, 5]}