Question

The radius of circle which passes through the focus of parabola $x^2=4y$ and touches it at point (6,9) is $k\sqrt{10}$, then $k=$

• A. 1

Answer 1

5

Answer 2

The parabola is $x^2 = 4y$, so $a=1$ and the focus is $F(0, 1)$. The point of tangency is $P(6, 9)$. The slope of the tangent at $P$ is $\frac{dy}{dx} = \frac{x}{2} = \frac{6}{2} = 3$. The slope of the normal at $P$ is $-\frac{1}{3}$. The equation of the normal is $y - 9 = -\frac{1}{3}(x - 6)$, which simplifies to $x + 3y = 33$. Let the center of the circle be $O(h, k')$. It lies on the normal, so $h + 3k' = 33$. The radius squared is the distance from $O$ to $F$ and from $O$ to $P$. $R^2 = (h-0)^2 + (k'-1)^2 = (h-6)^2 + (k'-9)^2$ $h^2 + k'^2 - 2k' + 1 = h^2 - 12h + 36 + k'^2 - 18k' + 81$ $-2k' + 1 = -12h - 18k' + 117$ $12h + 16k' = 116 \implies 3h + 4k' = 29$. Solving $h + 3k' = 33$ and $3h + 4k' = 29$: $h = 33 - 3k'$ $3(33 - 3k') + 4k' = 29$ $99 - 9k' + 4k' = 29 \implies 5k' = 70 \implies k' = 14$. $h = 33 - 3(14) = 33 - 42 = -9$. Center is $O(-9, 14)$. Radius squared $R^2 = (-9-0)^2 + (14-1)^2 = 81 + 169 = 250$. $R = \sqrt{250} = 5\sqrt{10}$. Given $R = k\sqrt{10}$, so $k=5$.