1971 AMC 12 第 19 题

Question

若直线

y=mx+1

与椭圆

x^2+4y^2=1

恰有一个交点，则

m^2

的值为：

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}

将直线方程代入椭圆方程

Substitute the line equation into the ellipse

恰有一个交点意味着所得二次方程的判别式为零

Exactly one intersection means the resulting quadratic has discriminant zero

代入可得

(1+4m^2)x^2+8mx+3=0\text{。}

相切要求判别式为零：

64m^2-12(1+4m^2)=0\text{。}

因此

16m^2=12

，所以

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

。

因此，正确答案为 C。

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,

so

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

Therefore, the correct answer is C.

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