1971 AMC 12 Problem 19

Attempt Problem 19 of the 1971 AMC 12 below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 1971 AMC 12 solutions, or check the answer key.

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19.

If the line

y=mx+1

intersects the ellipse

x^2+4y^2=1

exactly once, then the value of

m^2

is:

\frac{1}{2}

\frac{2}{3}

\frac{3}{4}

\frac{4}{5}

\frac{5}{6}

Substitute the line equation into the ellipse

Exactly one intersection means the resulting quadratic has discriminant zero

Substitution gives

(1+4m^2)x^2+8mx+3=0.

Tangency requires its discriminant to vanish:

64m^2-12(1+4m^2)=0.

Thus

16m^2=12,

m^2=\frac{3}{4}.

Therefore, the correct answer is C.

Problem 19 in other years