Question

\((1,1)\) is the focus of the parabola \[ y^{2}-4ax-2ay+a^{2}=0. \] If the circles \[ (x-\alpha)^2+(y-\beta)^2=r^2 \] touch the X-axis and the axis of the given parabola, then \[ \{(\alpha,\beta)\} \] is:

• A. a line \(y=\frac12\)
• B. a line \(y=1\)
• C. a circle \(x^2+y^2=\frac14\)
• D. a parabola \(y^2=2x\)

Hint:

For a circle tangent to two lines, equate the perpendicular distances from the centre to both lines.

Answer 1

The Correct Option is A

Concept: The axis of a parabola is the line passing through its vertex and focus. A circle touching two perpendicular lines has centre equidistant from both lines.

Step 1: Find the parabola axis.
Given \[ y^2-4ax-2ay+a^2=0. \] Complete square: \[ (y-a)^2=4ax. \] This is a parabola with vertex \[ (0,a). \] Focus \[ (a,a). \] Given focus is \[ (1,1). \] Thus \[ a=1. \] Hence parabola becomes \[ (y-1)^2=4x. \] Its axis is \[ y=1. \]

Step 2: Use touching conditions.
Circle touches X-axis. Distance of centre from X-axis \[ = |\beta| = r. \] Circle also touches axis \[ y=1. \] Distance from centre to axis \[ = |\beta-1| = r. \] Thus \[ |\beta| = |\beta-1|. \]

Step 3: Solve for \(\beta\).
\[ \beta^2=(\beta-1)^2 \] \[ \beta^2=\beta^2-2\beta+1 \] \[ 2\beta=1 \] \[ \beta=\frac12. \] Therefore all centres lie on a fixed horizontal line. Hence locus is \[ \boxed{y=\frac12}. \]