Question

The axis of symmetry of a parabola is $x=6$, and one point on the parabola is $(2,8)$. What is another point on the parabola?

Answer 1

There could be infinite parabolas , but along $(2,8)$ we have $(10,8)$ too on all parabolas as $x=6$ is axis of symmetry.

Explanation :

Given axis of symmetry $x=6$ , the equation of parabola is of the form $y=a(x-6{)}^2+k$

As it passes through $(2,8)$ we have

$8=a(2-6{)}^2+k$ or $16a=8-k$ or $a=\frac{8-k}{16}=\frac{1}{2}-\frac{k}{16}$

Hence, we could have a series of parabolas using different values of $k$. Let us choose $k=-8,24,40$, which gives $a=1,-1,-2$ and three equations are

$y=(x-6{)}^2-8$.

$y=-(x-6{)}^2+24$ and $y=-2(x-6{)}^2+40$

and parabolas appear as (graph not drawn to scale) follows.

graph$\left\{\right(y-(x-6{)}^2+8)(y+2(x-6{)}^2-40\left)\right(y+(x-6{)}^2-24)=0[0,15,-15.84,24.16]\}$

Note that it also passes through $(10,8)$ as $x=6$ is axis of symmetry.