$OA$ is the chord of the parabola $y^2 = 4x$ (where $O$ is the origin). $AB$ is a chord of $y^2 = 4x$ and perpendicular to $OA$ which cuts the axis of the parabola at $C$. If the foot of $A$ on the axis of the parabola is $D$, then the length $CD$ is equal to
Step-by-Step Solution
Key Concept: The normal to a parabola at a point has slope equal to the negative reciprocal of the tangent slope, and intersects the axis at specific coordinates derived from the normal equation.
The equation of the required normal is $y = -5x + 10a + 125a$, which simplifies to $y + 5x - 135a = 0$. From the given parabola $y^2 = -4x + 2at_1 + at_2^2$, we identify key points: $A$ is at $(t^2, 2t)$, and the slope of $OA$ is $\frac{2}{t}$. The slope of the normal $AB$ is $-\frac{t}{2}$. Using the normal equation with point $A(t^2, 2t)$, we derive $C = (4 + t^2, 0)$ and $D = (t^2, 0)$. Therefore, $CD = |t^2 + 4 - t^2| = 4$ units.
Correct Answer: 4