Question

Let \(S\) and \(S'\) be the foci of the ellipse \[ \frac{x^2}{25} + \frac{y^2}{9} = 1 \] and \(P(\alpha,\beta)\) be a point on the ellipse in the first quadrant. If \[ (SP)^2 + (S'P)^2 - SP \cdot S'P = 37, \] then \(\alpha^2 + \beta^2\) is equal to

• A. 17
• B. 13
• C. 15
• D. 11

Hint:

For ellipse problems involving foci, always use standard identities like \(SP + S'P = 2a\).

Answer 1

The Correct Option is B

The given ellipse is \[ \frac{x^2}{25} + \frac{y^2}{9} = 1, \] where \[ a^2 = 25,\quad b^2 = 9. \]
Step 1: Find the foci.
For an ellipse, \[ c^2 = a^2 - b^2 = 25 - 9 = 16 \Rightarrow c = 4. \] Hence, the foci are \[ S(4,0), \quad S'(-4,0). \]
Step 2: Use known distance relations for ellipse.
For any point \(P\) on the ellipse, \[ SP + S'P = 2a = 10. \]
Step 3: Evaluate the given expression.
Using the identity, \[ (SP)^2 + (S'P)^2 - SP \cdot S'P = (SP + S'P)^2 - 3(SP)(S'P), \] \[ = 10^2 - 3(SP)(S'P). \] Given this equals 37, \[ 100 - 3(SP)(S'P) = 37 \] \[ (SP)(S'P) = 21. \]
Step 4: Express \(SP \cdot S'P\) in terms of coordinates.
For a point \(P(\alpha,\beta)\), \[ SP \cdot S'P = \sqrt{(\alpha-4)^2 + \beta^2}\sqrt{(\alpha+4)^2 + \beta^2}. \] Squaring both sides, \[ (SP \cdot S'P)^2 = \left[(\alpha^2+\beta^2+16)^2 - (8\alpha)^2\right]. \] Using \(SP \cdot S'P = 21\), \[ (\alpha^2+\beta^2+16)^2 - 64\alpha^2 = 441. \]
Step 5: Use ellipse equation.
From the ellipse, \[ \frac{\alpha^2}{25} + \frac{\beta^2}{9} = 1 \Rightarrow 9\alpha^2 + 25\beta^2 = 225. \] Solving simultaneously gives \[ \alpha^2 + \beta^2 = 13. \]
Final Answer: \[ \boxed{13} \]