Question

If the focal chord drawn through the point $ P(5, 5) $ to the parabola $ y^2 = 5x $ meets the parabola again at the point $ Q $, then the tangent drawn to this parabola at $ Q $ meets the axis of the parabola at the point:

• A. \( \left( -\frac{5}{4}, 0 \right) \)
• B. \( \left( \frac{5}{16}, 0 \right) \)
• C. \( \left( -\frac{5}{16}, 0 \right) \)
• D. \( \left( \frac{5}{4}, 0 \right) \)

Hint:

For a parabola \( y^2 = 4ax \), the tangent at \( (x_1, y_1) \) is \( y y_1 = 2a (x + x_1) \). Focal chords pass through the focus.

Answer 1

The Correct Option is C

Step 1: Identify the parabola and focus.
Parabola: \( y^2 = 5x \), so \( a = \frac{5}{4} \). Focus: \( \left( \frac{5}{4}, 0 \right) \). Axis: \( y = 0 \).
Step 2: Find the focal chord through \( P(5, 5) \).
Line through focus and \( P \): Slope = \( \frac{4}{3} \). Equation: \( y = \frac{4}{3} \left( x - \frac{5}{4} \right) \).
Step 3: Find point \( Q \).
Substitute into parabola: \( \left( \frac{4}{3} \left( x - \frac{5}{4} \right) \right)^2 = 5x \). Solve:
\[ 16x^2 - 85x + 25 = 0 \quad \Rightarrow \quad x = 5, \quad x = \frac{5}{16}. \] \[ Q = \left( \frac{5}{16}, -\frac{5}{4} \right). \] Step 4: Tangent at \( Q \).
Tangent: \( y \left( -\frac{5}{4} \right) = \frac{5}{2} \left( x + \frac{5}{16} \right) \). Intersects x-axis at \( \left( -\frac{5}{16}, 0 \right) \).