Question

A circle is drawn with its centre at the focus of the parabola \( y^2 = 2px \) such that it touches the directrix of the parabola. Then a point of intersection of the circle and the parabola is

• A. \( (2p, 2p) \)
• B. \( \left( \frac{p}{2}, -p \right) \)
• C. \( (2p, -2p) \)
• D. \( \left( p, \sqrt{2p} \right) \)

Hint:

When a circle is centered at the focus of a parabola and touches the directrix, its radius equals the distance from focus to directrix. Use this to form the equation of the circle.

Answer 1

The Correct Option is B

Step 1: Geometry of the parabola.
For the parabola \( y^2 = 2px \), the focus is at \( \left( \frac{p}{2}, 0 \right) \), and the directrix is the line \( x = -\frac{p}{2} \). Step 2: Geometry of the circle.
The circle is centered at the focus and touches the directrix. Therefore, its radius is the distance from the focus to the directrix, which is: \[ r = \left| \frac{p}{2} - \left( -\frac{p}{2} \right) \right| = p \] So the equation of the circle is: \[ (x - \frac{p}{2})^2 + y^2 = p^2 \] Step 3: Solve the system.
Substitute \( y^2 = 2px \) into the circle's equation: \[ (x - \frac{p}{2})^2 + 2px = p^2 \] Expand: \[ x^2 - px + \frac{p^2}{4} + 2px = p^2 \Rightarrow x^2 + px + \frac{p^2}{4} = p^2 \Rightarrow x^2 + px - \frac{3p^2}{4} = 0 \] Solve this quadratic in \( x \). One root is \( x = \frac{p}{2} \). Using \( y^2 = 2px \), we get: \[ y^2 = 2p . \frac{p}{2} = p^2 \Rightarrow y = \pm p \] Thus, one point is \( \left( \frac{p}{2}, -p \right) \)