Question

Let S be the focus of the parabola \( y^{2}=36x \) and let the line \( x+by+c=0 \) intersect the parabola at the points Q and R. If the centroid of \( \triangle QRS \) is (57,0), then -c is

• A. an even number
• B. a prime number
• C. a perfect square
• D. a perfect cube

Hint:

Whenever the y-coordinate of the centroid of a triangle inscribed in a horizontal parabola is zero, the chord connecting the other two vertices must be a perfectly vertical line, meaning \( b = 0 \) automatically!

Answer 1

The Correct Option is C

Step 1: Find the focus.
Compare $y^2=36x$ with $y^2=4ax$: $4a=36$, so $a=9$. The focus is $S(9,0)$.
Step 2: Name the two points on the parabola.
Use parametric points $Q(9t_1^2,\,18t_1)$ and $R(9t_2^2,\,18t_2)$, since a general point is $(at^2,2at)$ with $a=9$.
Step 3: Use the $y$-coordinate of the centroid.
The centroid of $\triangle QRS$ is $(57,0)$. For the $y$-part: \[ \frac{18t_1+18t_2+0}{3}=0\implies t_1+t_2=0\implies t_2=-t_1 \]
Step 4: Use the $x$-coordinate of the centroid.
\[ \frac{9t_1^2+9t_2^2+9}{3}=57\implies t_1^2+t_2^2+1=19 \] Since $t_2=-t_1$, this is $2t_1^2+1=19$, so $t_1^2=9$ and $t_1=3,\ t_2=-3$.
Step 5: Find the line through $Q$ and $R$.
$Q=(81,54)$ and $R=(81,-54)$ share $x=81$, so the line is $x=81$, i.e. $x-81=0$. Comparing with $x+by+c=0$ gives $c=-81$.
Step 6: Identify $-c$.
\[ -c=81=9^2 \] So $-c$ is a perfect square. \[ \boxed{\text{a perfect square}} \]