Question

If a point $ P(x, y) $ satisfies the condition that its distance from the point $ (3, -2) $ is equal to its distance from the line $ y = 2x + 1 $, then the locus of point $ P $ is:

• A. A parabola
• B. A circle
• C. A straight line
• D. A pair of straight lines

Hint:

The definition of a parabola is key: it is the set of points equidistant from a point and a line.

Answer 1

The Correct Option is A

To find the locus of point \( P(x, y) \), we set its distance from \( (3, -2) \) equal to its distance from the line \( y = 2x + 1 \).

The distance from \( P(x, y) \) to \( (3, -2) \) is \( d_1 = \sqrt{(x - 3)^2 + (y + 2)^2} \).

The distance from \( P(x, y) \) to \( y = 2x + 1 \) is \( d_2 = \frac{|y - 2x - 1|}{\sqrt{2^2+1^2}} = \frac{|y - 2x - 1|}{\sqrt{5}} \).

Equating the distances, \( d_1 = d_2 \):

\[ \sqrt{(x - 3)^2 + (y + 2)^2} = \frac{|y - 2x - 1|}{\sqrt{5}} \]

Squaring both sides yields:

\[ 5[(x - 3)^2 + (y + 2)^2] = (y - 2x - 1)^2 \]

Expanding and simplifying gives:

\[ 5(x^2 - 6x + 9 + y^2 + 4y + 4) = y^2 - 4xy + 4x^2 - 2y + 4x + 1 \]

\[ 5x^2 - 30x + 45 + 5y^2 + 20y + 20 = y^2 - 4xy + 4x^2 - 2y + 4x + 1 \]

Rearranging terms to one side:

\[ 5x^2 - 4x^2 + 5y^2 - y^2 + 4xy - 30x - 4x + 20y + 2y + 45 + 20 - 1 = 0 \]

Simplifying results in:

\[ x^2 + 4xy + 4y^2 - 34x + 22y + 64 = 0 \]

This equation represents a parabola.

Therefore, the locus of point \( P \) is a parabola.