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: \[ \frac{(y - 0)^2}{16} + \frac{(x - 2)^2}{7} = 1 \Rightarrow 7y^2 + 16(x^2 - 4x + 4) = 112 \Rightarrow 16x^2 + 7y^2 - 64x - 48 = 0 \]

Answer 2

Step 1: Understand the problem
Given two fixed points \( A(2,3) \) and \( B(2,-3) \), we need to find the equation of the locus of point \( P(x,y) \) such that the sum of distances from \( P \) to \( A \) and \( B \) is constant:
\[ PA + PB = 8 \]
This defines an ellipse with foci at \( A \) and \( B \).

Step 2: Write the distance expressions
\[ PA = \sqrt{(x - 2)^2 + (y - 3)^2}, \quad PB = \sqrt{(x - 2)^2 + (y + 3)^2} \]

Step 3: Use the ellipse property
The sum of distances to foci is constant:
\[ \sqrt{(x - 2)^2 + (y - 3)^2} + \sqrt{(x - 2)^2 + (y + 3)^2} = 8 \]

Step 4: Simplify by substitution
Let \( u = x - 2 \), so:
\[ \sqrt{u^2 + (y - 3)^2} + \sqrt{u^2 + (y + 3)^2} = 8 \]

Step 5: Square both sides
Set \( S_1 = \sqrt{u^2 + (y - 3)^2} \), \( S_2 = \sqrt{u^2 + (y + 3)^2} \).
\[ S_1 + S_2 = 8 \implies (S_1 + S_2)^2 = 64 \]
\[ S_1^2 + 2 S_1 S_2 + S_2^2 = 64 \]

Step 6: Substitute \( S_1^2 \) and \( S_2^2 \)
\[ [u^2 + (y - 3)^2] + 2 S_1 S_2 + [u^2 + (y + 3)^2] = 64 \]
\[ 2 u^2 + (y - 3)^2 + (y + 3)^2 + 2 S_1 S_2 = 64 \]
Expand squares:
\[ (y - 3)^2 = y^2 - 6y + 9, \quad (y + 3)^2 = y^2 + 6y + 9 \]
Sum:
\[ (y - 3)^2 + (y + 3)^2 = 2 y^2 + 18 \]

Step 7: Update equation
\[ 2 u^2 + 2 y^2 + 18 + 2 S_1 S_2 = 64 \implies 2 S_1 S_2 = 64 - 18 - 2 u^2 - 2 y^2 = 46 - 2 u^2 - 2 y^2 \]
\[ S_1 S_2 = 23 - u^2 - y^2 \]

Step 8: Express \( S_1 S_2 \) explicitly
\[ S_1 S_2 = \sqrt{u^2 + (y - 3)^2} \times \sqrt{u^2 + (y + 3)^2} = \sqrt{[u^2 + (y - 3)^2][u^2 + (y + 3)^2]} \]
Calculate inside:
\[ [u^2 + (y - 3)^2][u^2 + (y + 3)^2] = (u^2 + y^2 - 6y + 9)(u^2 + y^2 + 6y + 9) \]
Use \((a - b)(a + b) = a^2 - b^2\) with \(a = u^2 + y^2 + 9\) and \(b = 6y\):
\[ = (u^2 + y^2 + 9)^2 - (6 y)^2 = (u^2 + y^2 + 9)^2 - 36 y^2 \]

Step 9: Square both sides of \( S_1 S_2 = 23 - u^2 - y^2 \)
\[ (S_1 S_2)^2 = (23 - u^2 - y^2)^2 \]
So,
\[ (u^2 + y^2 + 9)^2 - 36 y^2 = (23 - u^2 - y^2)^2 \]

Step 10: Substitute back \( u = x - 2 \)
Rewrite:
\[ \big[(x - 2)^2 + y^2 + 9\big]^2 - 36 y^2 = \big[23 - (x - 2)^2 - y^2 \big]^2 \]

Step 11: Expand and simplify
Expanding and simplifying leads to the ellipse equation:
\[ 16 x^2 + 7 y^2 - 64 x - 48 = 0 \]

Final answer: The equation of the locus is
\[ 16 x^2 + 7 y^2 - 64 x - 48 = 0 \]