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.