Problem 503

The equation of a parabola, written in the form $y=(x-p)^2+q$, has its vertex at $(p,q)$.

Thus, the parabola with equation $y=(x-3)^2+1$, has its vertex at $(3,1)$.
Solution 1

Under a translation, the shape of a parabola remains unchanged.

That is, the new parabola is congruent to the original parabola.

After a translation of 3 units to the left and 3 units up, the original vertex $(3,1)$ moves to the point $(3-3,1+3)$ or $(0,4)$.

Since this new parabola is congruent to the original, but it has its vertex at $(0,4)$, then its equation is $y=(x-0)^2+4$ or $y=x^2+4$.

Solution 2

Under a translation of 3 units left and 3 units up, the equation $y=(x-3)^2+1$ becomes $$\begin{aligned} y-3& =((x+3)-3)^2+1\\ y-3& =x^2+1\\ y& =x^2+4\end{aligned}$$
At the point of intersection of these two parabolas, their $y$ values must be equal. Thus, $$\begin{aligned} (x-3)^2+1& =x^2+4\\ x^2-6x+9+1& =x^2+4\\ -6x& =-6\\ x& =1\end{aligned}$$ Substituting $x=1$ into the equation $y=x^2+4$, we determine the $y$ value of the point of intersection to be $y=5$.

Therefore, the two parabolas intersect at the point $(1,5)$.
At the point of intersection of these two parabolas, their $y$ values must be equal. Thus, $$\begin{aligned} (x-3)^2+1& =ax^2+4\\ x^2-6x+9+1& =ax^2+4\\ 0& =ax^2-x^2+6x-6\\ 0& =(a-1)x^2+6x-6\\\end{aligned}$$ Since the two parabolas intersect at exactly one point, then the resulting equation $(a-1)x^2+6x-6=0$ (which is quadratic since $a<0$), has exactly one solution.

Thus, the discriminant of this equation must equal zero.

(Note: The discriminant of a quadratic equation of the form $ax^2+bx+c=0$, is $b^2-4ac$.)

Solving $6^2-4(-6)(a-1)=0$, we get $36+24(a-1)=0$ or $24a=-12$, and so $a=-\frac12$.

That is, the parabolas with equations $y=ax^2+4$ and $y=(x-3)^2+1$ touch at exactly one point when $a=-\frac12$.