Question

Consider the circle x² + y² = 9 and the parabola y² = 8x. They intersect at P and Q in the first and the fourth quadrants, respectively. Tangents to the circle at P and Q intersect the x-axis at R and tangents to the parabola at P and Q intersect the x-axis at S.

• A. 1: √2
• B. 1:2
• C. 1:4
• D. 1:8

Answer 1

Answer: (C)

1:4

Answer 2

The intersection points are $P(1, 2\sqrt{2})$ and $Q(1, -2\sqrt{2})$. The tangent to the circle $x^2 + y^2 = 9$ at $P(1, 2\sqrt{2})$ is $x + 2\sqrt{2}y = 9$. Setting $y=0$, we get $x=9$, so $R=(9,0)$. The tangent to the parabola $y^2 = 8x$ at $P(1, 2\sqrt{2})$ is $y(2\sqrt{2}) = 2(2)(x+1)$, which simplifies to $2\sqrt{2}y = 4(x+1)$. Setting $y=0$, we get $x=-1$, so $S=(-1,0)$.

The base $PQ$ is common to both triangles $PQS$ and $PQR$. The length of $PQ$ is $|2\sqrt{2} - (-2\sqrt{2})| = 4\sqrt{2}$. The height of $\triangle PQS$ with respect to base $PQ$ is the perpendicular distance from $S(-1,0)$ to the line $x=1$, which is $|1 - (-1)| = 2$. Area($\triangle PQS$) = $\frac{1}{2} \times 4\sqrt{2} \times 2 = 4\sqrt{2}$. The height of $\triangle PQR$ with respect to base $PQ$ is the perpendicular distance from $R(9,0)$ to the line $x=1$, which is $|9 - 1| = 8$. Area($\triangle PQR$) = $\frac{1}{2} \times 4\sqrt{2} \times 8 = 16\sqrt{2}$. The ratio of the areas is $\frac{\text{Area}(\triangle PQS)}{\text{Area}(\triangle PQR)} = \frac{4\sqrt{2}}{16\sqrt{2}} = \frac{1}{4}$.