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Algebra Difficulty 3.4 AMC 10/12 Find the answer

In the Cartesian coordinate system xoyxoy, the parametric equation of curve CC is {x=8tan2θy=8tanθ\begin{cases} x=8\tan^{2}\theta \\ y=8\tan\theta \end{cases} (where θ\theta is the parameter, θ(π2,π2)\theta \in \left(-\frac{\pi}{2}, \frac{\pi}{2}\right)). In the polar coordinate system with OO as the pole and the positive half-axis of xx as the polar axis, the equation of line ll is ρcos(θπ4)=42.\rho\cos\left(\theta - \frac{\pi}{4}\right) = -4\sqrt{2}.
(1) Find the Cartesian coordinate equation of line ll;
(2) If PP is a point on curve CC and QQ is a point on line ll, find the minimum value of PQ|PQ|.

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

Solution

Solution:
(1) Since the equation of line ll is ρcos(θπ4)=42,\rho\cos\left(\theta - \frac{\pi}{4}\right) = -4\sqrt{2},
it follows that ρcosθcosπ4+ρsinθsinπ4=42,\rho\cos\theta\cos\frac{\pi}{4} + \rho\sin\theta\sin\frac{\pi}{4} = -4\sqrt{2},
thus, the Cartesian coordinate equation of line ll is 22x+22y=42,\frac{\sqrt{2}}{2}x + \frac{\sqrt{2}}{2}y = -4\sqrt{2}, which simplifies to x+y+8=0x + y + 8 = 0.
(2) Since the parametric equation of curve CC is {x=8tan2θy=8tanθ\begin{cases} x=8\tan^{2}\theta \\ y=8\tan\theta \end{cases} (where θ\theta is the parameter, θ(π2,π2)\theta \in \left(-\frac{\pi}{2}, \frac{\pi}{2}\right)),
and PP is a point on curve CC and QQ is a point on line ll,
the distance dd from point P(8tan2θ,8tanθ)P(8\tan^2\theta, 8\tan\theta) to line ll is:
d=8tan2θ+8tanθ+82=42(tanθ+12)2+34=42(tanθ+12)2+32,d = \frac{|8\tan^{2}\theta + 8\tan\theta + 8|}{\sqrt{2}} = 4\sqrt{2}\left|\left(\tan\theta + \frac{1}{2}\right)^2 + \frac{3}{4}\right| = 4\sqrt{2}\left(\tan\theta + \frac{1}{2}\right)^2 + 3\sqrt{2},
thus, when tanθ=12\tan\theta = -\frac{1}{2}, PQ|PQ| reaches its minimum value of 323\sqrt{2}.
Therefore, the minimum value of PQ|PQ| is 32\boxed{3\sqrt{2}}.

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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.