Competition problem — Algebra — Maths Olympiad Prep

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Algebra Difficulty 3.5 AMC 10/12 Find the answer

In the Cartesian coordinate system $(xOy)$ , let $l$ be a line with an angle of inclination $\alpha$ and parametric equations $\begin{cases} x = 3 + t\cos\alpha \\ y = t\sin\alpha \end{cases}$ (where $t$ is a parameter). The line $l$ intersects the curve $C$ : $\begin{cases} x = \frac{1}{\cos\theta} \\ y = \tan\theta \end{cases}$ (where $\theta$ is a parameter) at two distinct points $A$ and $B$ .

1. If $\alpha = \frac{\pi}{3}$ , find the rectangular coordinates of the midpoint of the line segment $AB$ .
2. If the slope of line $l$ is 2 and it passes through the known point $P(3, 0)$ , find the value of $|PA| \cdot |PB|$ .

Your answer

A number or a short expression. Spacing and $ signs are ignored.

Solution

1. From the curve $C$ : $\begin{cases} x = \frac{1}{\cos\theta} \\ y = \tan\theta \end{cases}$ , we can derive the standard equation of $C$ as $x^2 - y^2 = 1$ .

When $\alpha = \frac{\pi}{3}$ , the parametric equations of line $l$ become $\begin{cases} x = 3 + \frac{1}{2}t \\ y = \frac{\sqrt{3}}{2}t \end{cases}$ .

Substituting these parametric equations into the standard equation of curve $C$ , we get $t^2 - 6t - 16 = 0$ .

The midpoint of the line segment $AB$ corresponds to the value $t = 3$ .

Hence, the rectangular coordinates of the midpoint of line segment $AB$ are $(\frac{9}{2}, \frac{3\sqrt{3}}{2})$ .

2. Substitute the parametric equations of line $l$ into the standard equation of curve $C$ , and simplify to get $(\cos^2\alpha - \sin^2\alpha)tx^2 + 6\cos\alpha t + 8 = 0$ .

Then, $|PA| \cdot |PB| = |\frac{8}{\cos^2\alpha - \sin^2\alpha}| = |\frac{8(1 + \tan^2\alpha)}{1 - \tan^2\alpha}| = \boxed{\frac{40}{3}}$ .

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Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic and difficulty added by this site.