1966 AHSME Problems/Problem 40

Problem

[asy] draw(circle((0,0),10),black+linewidth(1)); MP("O", (0,0), S);MP("A", (-10,0), W);MP("B", (10,0), E);MP("C", (10,10), E);MP("D", (6,8), N); MP("a", (-5,0), S);MP("E", (-6,3), N); dot((0,0));dot((-6,2)); draw((-10,0)--(10,0),black+linewidth(1)); draw((-10,0)--(10,10),black+linewidth(1)); draw((-10,-12)--(-10,12),black+linewidth(1)); draw((10,-12)--(10,12),black+linewidth(1)); [/asy]

In this figure $AB$ is a diameter of a circle, centered at $O$, with radius $a$. A chord $AD$ is drawn and extended to meet the tangent to the circle at $B$ in point $C$. Point $E$ is taken on $AC$ so the $AE=DC$. Denoting the distances of $E$ from the tangent through $A$ and from the diameter $AB$ by $x$ and $y$, respectively, we can deduce the relation:

$\text{(A) } y^2=\frac{x^3}{2a-x} \quad \text{(B) } y^2=\frac{x^3}{2a+x} \quad \text{(C) } y^4=\frac{x^2}{2a-x} \\ \text{(D) } x^2=\frac{y^2}{2a-x} \quad \text{(E) } x^2=\frac{y^2}{2a+x}$

Solution

$\fbox{A}$

See also

1966 AHSME (ProblemsAnswer KeyResources)
Preceded by
Problem 39
Followed by
Problem 40
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