Track / Stage 3 / 75 of 260 #75 of 2000

Problem 75

AMC 10/12, early questions

Geometry Difficulty 3.3 Find the answer

Given a parabola $M$ : $y^2=2px$ ( $p > 0$ ) and a circle: $x^2+(y-4)^2=a^2$ , the common point in the first quadrant is $A$ . The distance from point $A$ to the focus of parabola $M$ is $a$ . The sum of the distance from a point on parabola $M$ to its directrix and the distance to point $C$ has a maximum value of $2$ . $O$ is the origin, and the length of the chord intercepted by line $OA$ on circle $C$ is ( $\quad\quad$ ).

A: $2$
B: $2 \sqrt {3}$
C: $\dfrac {7 \sqrt {2}}{3}$
D: $\dfrac {7 \sqrt {2}}{6}$

Your answer

Official solution

The circle $C$ : $x^2+(y-4)^2=a^2$ has center $C(0,4)$ and radius $a$ .

According to the given conditions, we have $a = \dfrac {p}{4}+ \dfrac {p}{2}= \dfrac {3 \sqrt {2}}{2}$ , so $A( \dfrac { \sqrt {2}}{2}, \dfrac {3 \sqrt {2}}{2})$ .

The distance $d$ from point $C$ to line $OA$ : $y=2 \sqrt {2}x$ is $d= \dfrac {|0-4|}{ \sqrt {8+1}}= \dfrac {4}{3}$ .

Thus, the length of the chord intercepted by line $OA$ on circle $C$ is $2 \sqrt {( \dfrac {3 \sqrt {2}}{2})^{2}-( \dfrac {4}{3})^{2}}= \boxed{\dfrac {7 \sqrt {2}}{3}}$ .

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