Track / Stage 3 / 89 of 260 #89 of 2000

Problem 89

AMC 10/12, early questions

Algebra Difficulty 3.2 Find the answer

In the rectangular coordinate system, a polar coordinate system is established with the origin as the pole and the positive semi-axis of the $x$ -axis as the polar axis. Given the curve $C:ρ\sin^2θ=2a\cos θ (a > 0)$ , the line $l:\begin{cases}x=-2+\frac{\sqrt{2}}{2}t\\y=-4+\frac{\sqrt{2}}{2}t\end{cases} (t$ is the parameter $)$ passes through point $P(-2,-4)$ and intersects curve $C$ at points $M$ and $N$ .

(I) Find the ordinary equations of curve $C$ and line $l$ ;

(II) If $|PM|, |MN|, |PN|$ form a geometric sequence, find the value of real number $a$ .

Your answer

Official solution

(I) Since $\begin{cases} x=ρ\cos θ \\ y=ρ\sin θ \end{cases}$ , multiplying both sides of the equation $ρ\sin^2θ=2a\cos θ (a > 0)$ by $ρ$ , we get the rectangular coordinate equation of curve $C$ as $y^2=2ax (a > 0)$ . The ordinary equation of line $l$ is $x-y-2=0$ .

(II) Solving the system of equations $\begin{cases} y^2=2ax \\ x-y-2=0 \end{cases}$ , we eliminate $y$ and obtain $t^2-2(4+a)\sqrt{2}t+8(4+a)=0 \; (\*).$

Let points $M$ and $N$ correspond to parameters $t_1$ and $t_2$ , which are the roots of the above equation. Then, we have $|PM|=|t_1|$ , $|PN|=|t_2|$ , and $|MN|=|t_1-t_2|$ .

According to the problem, we have $(t_1-t_2)^2=|t_1t_2|$ , which implies $(t_1+t_2)^2-4t_1t_2=|t_1t_2|$ .

From $(\*),$ we have $t_1+t_2=2(4+a)\sqrt{2}$ and $t_1t_2=8(4+a) > 0$ .

This leads to $(4+a)^2-5(4+a)=0$ , which gives $a=1$ or $a=-4$ .

Since $a > 0$ , we have $\boxed{a=1}$ .

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