Question

Find the equation of the parabola whose axis is parallel to $y$-axis which passes through points $(4,5)$,$(-2,11)$ and $(-4,21)$.

Answer 1

A parabola with axis parallel to y-axis has equal to $y=a{x}^2+bx+c$, for constants a, b, c in each of the three points.

$5=a(4{)}^2+b(4)+c$

$11=a(-2{)}^2+b(-2)+c$

$21=a(-4{)}^2+b(-4)+c$

So, $\left(\begin{array}{ccc}16& 4& 1\\ 4& -2& 1\\ 16& -4& 1\end{array}\right|\begin{array}{c}5\\ 11\\ 21\end{array})$ is our system.

This simplifies as follows

$\Rightarrow \left(\begin{array}{ccc}16& 4& 1\\ 4& -2& 1\\ 0& -8& 0\end{array}\right|\begin{array}{c}5\\ 11\\ 16\end{array})$

$\Rightarrow \left(\begin{array}{ccc}1& 1/4& 1/16\\ 0& -3& 3/4\\ 0& 1& 0\end{array}\right|\begin{array}{c}5/16\\ 39/4\\ -2\end{array})$

$\Rightarrow \left(\begin{array}{ccc}1& 0& 1/16\\ 0& 1& 0\\ 0& 0& 1\end{array}\right|\begin{array}{c}13/16\\ -2\\ 5\end{array})$

Now we get

$c=5$

$b=2$

$a=\frac{13}{16}-5\frac{1}{16}$

$=\frac{13-5}{16}$

$=\frac{8}{16}$

$=\frac{1}{2}$

Finally we write $y=\frac{1}{2}.x^2-2x+5$.