Question

Let $T$ be the tangent to the parabola $y^2 = 16x$ at the point $(64, 32)$. Let $L$ be the tangent to the same parabola at another point $(x_1, y_1)$ on the parabola. If $L$ and $T$ are perpendicular to each other, then the distance between the point $(x_1, y_1)$ and the focus of the parabola, is

• A. $\frac{15}{4}$
• B. 4
• C. $\frac{17}{4}$
• D. 5

Hint:

A fundamental property of parabolas is that the point of intersection of two perpendicular tangents lies on the directrix. Additionally, focal distance is simply $x + a$ because the distance to focus equals the distance to the directrix.

Answer 1

The Correct Option is C

Step 1: Understanding the Question:
This coordinate geometry problem requires finding the focal distance of a specific point on a parabola. The point is defined by its tangent being perpendicular to another given tangent.
Step 2: Key Formula or Approach:

Slope of tangent to $y^2 = 4ax$ at $(x_0, y_0)$ is $m = \frac{2a}{y_0}$.

For perpendicular lines, $m_1 \cdot m_2 = -1$.

Focal distance of point $(x_1, y_1)$ on parabola $y^2 = 4ax$ is $a + x_1$.

Step 3: Detailed Explanation:

Parabola is $y^2 = 16x$, so $4a = 16 \implies a = 4$. Focus is at $(4, 0)$.

Slope of tangent $T$ at $(64, 32)$: \[ m_T = \frac{2a}{y_0} = \frac{8}{32} = \frac{1}{4} \]
Since $L$ is perpendicular to $T$, the slope of $L$ is $m_L = -4$.

For a tangent with slope $m$, the point of contact is $(x_1, y_1) = \left( \frac{a}{m^2}, \frac{2a}{m} \right)$.

Substituting $a = 4$ and $m = -4$: \[ x_1 = \frac{4}{(-4)^2} = \frac{4}{16} = \frac{1}{4} \] \[ y_1 = \frac{2(4)}{-4} = -2 \]
The point $(x_1, y_1)$ is $\left( \frac{1}{4}, -2 \right)$.

The distance of any point $(x_1, y_1)$ on the parabola from the focus is given by the formula $d = a + x_1$.
\[ d = 4 + \frac{1}{4} = \frac{17}{4} \]
Step 4: Final Answer:
The distance is $\frac{17}{4}$.