1971 AHSME Problems/Problem 19


If the line $y=mx+1$ intersects the ellipse $x^2+4y^2=1$ exactly once, then the value of $m^2$ is

$\textbf{(A) }\textstyle\frac{1}{2}\qquad \textbf{(B) }\frac{2}{3}\qquad \textbf{(C) }\frac{3}{4}\qquad \textbf{(D) }\frac{4}{5}\qquad \textbf{(E) }\frac{5}{6}$


Plug in $y=mx+1$ into the ellipse's equation to find the intersection points: \[x^2 + 4(mx+1)^2 = 1\] After simplifying, we have a quadratic in $x$: \[(4m^2 + 1)x^2 + 8mx  +3 = 0\]

Because there is only one intersection point, then the quadratic has only one solution. This can only happen when the discriminant is 0. \[\Delta = b^2 - 4ac = (8m)^2 - (4)(1+4m^2)(3) = 0\]

Solving, we find $m^2 = \frac{3}{4}.$ The answer is $\textbf{(C)}.$

-edited by coolmath34