1966 AMC 12 Problem 40

Question

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 that

AE=DC.

If the coordinates of

E

are

(x,y),

then:

y^2=\dfrac{x^3}{2a-x}

y^2=\dfrac{x^3}{2a+x}

y^4=\dfrac{x^2}{2a-x}

x^2=\dfrac{y^2}{2a-x}

x^2=\dfrac{y^2}{2a+x}

Drop perpendiculars from

E

and

D

to diameter

AB

Use

AE=DC

between the two parallel tangents, then combine the altitude theorem with similar triangles

Drop perpendiculars

EM

and

DN

to

AB.

Since

AE=DC

and the tangents through

A,B

are parallel, their projections give

NB=x.

Hence

AN=2a-x.

In right triangle

ADB,

the altitude theorem gives

DN^2=x(2a-x).

Similar triangles

AME

and

AND

give

\frac{DN}{2a-x}=\frac{y}{x},

so

DN=\frac{y(2a-x)}{x}.

Substitution and cancellation yield

y^2=\frac{x^3}{2a-x}.

Therefore, the correct answer is A.

Problem 40 in other years