Question

If A(2,3) and B(2,-3) are two points, then the equation of the locus of a point $P$ such that $PA + PB = 8$ is

• A. $16x^2 + 7y^2 - 64x - 48 = 0$
• B. $16x^2 + 7y^2 - 64x + 48 = 0$
• C. $16x^2 - 7y^2 + 64x - 48 = 0$
• D. $16x^2 - 7y^2 + 64x + 48 = 0$

Hint:

Ellipse Geometry: For $PA + PB = \textconstant$, use ellipse properties: center at midpoint of foci, and $a^2 = b^2 + c^2$ helps get the full equation.

Answer 1

The Correct Option is A

$PA + PB = \text{constant}$ implies $P$ traces an ellipse with foci at $A$ and $B$. Distance between foci: \[ AB = \sqrt{(2 - 2)^2 + (3 + 3)^2} = 6 \Rightarrow 2c = 6 \Rightarrow c = 3 \] Given: $PA + PB = 8 = 2a \Rightarrow a = 4$ Using: $a^2 = b^2 + c^2 \Rightarrow 16 = b^2 + 9 \Rightarrow b^2 = 7$ Center = midpoint of A and B = $(2, 0)$, vertical major axis. Ellipse equation: \[ \frac{(y - 0)^2}{16} + \frac{(x - 2)^2}{7} = 1 \Rightarrow 7y^2 + 16(x^2 - 4x + 4) = 112 \Rightarrow 16x^2 + 7y^2 - 64x - 48 = 0 \]