Question

If S and S' are the foci of an ellipse \( \frac{x^2}{169} + \frac{y^2}{144} = 1 \) and the point B lying on positive Y-axis is one end of its minor axis, then the incentre of the triangle SBS' is

• A. \( \left(0, \frac{10}{3}\right) \)
• B. \( \left(\frac{13}{3}, \frac{10}{3}\right) \)
• C. \( \left(\frac{10}{3}, \frac{13}{3}\right) \)
• D. \( \left(0, \frac{13}{3}\right) \)

Hint:

For an isosceles triangle with vertices \( (\pm k, 0) \) and \( (0, h) \), the incenter always lies on the y-axis. You only need to calculate the y-coordinate.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:

We need to determine the coordinates of the foci \( S, S' \) and the vertex \( B \). The incenter of a triangle is found using the formula \( I = \left(\frac{ax_1+bx_2+cx_3}{a+b+c}, \frac{ay_1+by_2+cy_3}{a+b+c}\right) \), where \( a, b, c \) are the lengths of the sides opposite to vertices \( A, B, C \) respectively.
Step 2: Key Formula or Approach:

1. Eccentricity \( e = \sqrt{1 - \frac{b^2}{a^2}} \). 2. Foci coordinates \( (\pm ae, 0) \). 3. Incenter formula for \( \Delta SBS' \).
Step 3: Detailed Explanation:

Given ellipse: \( \frac{x^2}{13^2} + \frac{y^2}{12^2} = 1 \). Here \( a = 13, b = 12 \). Calculate eccentricity: \[ e = \sqrt{1 - \frac{144}{169}} = \sqrt{\frac{25}{169}} = \frac{5}{13} \] Foci \( S \) and \( S' \): \[ ae = 13 \times \frac{5}{13} = 5 \] So, \( S(5, 0) \) and \( S'(-5, 0) \). Point \( B \) on positive Y-axis is \( (0, b) = (0, 12) \). Calculate side lengths of \( \Delta SBS' \): 1. Side opposite \( B \) (base \( SS' \)): \( c = \sqrt{(5 - (-5))^2} = 10 \). 2. Side opposite \( S' \) (side \( SB \)): \( b = \sqrt{(5-0)^2 + (0-12)^2} = \sqrt{25+144} = 13 \). 3. Side opposite \( S \) (side \( S'B \)): \( a = \sqrt{(-5-0)^2 + (0-12)^2} = \sqrt{25+144} = 13 \). Calculate Incenter \( (x, y) \): Since the triangle is isosceles with the y-axis as the axis of symmetry, the x-coordinate of the incenter is 0. \[ y = \frac{a(y_{S}) + b(y_{S'}) + c(y_{B})}{a+b+c} \] \[ y = \frac{13(0) + 13(0) + 10(12)}{13+13+10} = \frac{120}{36} = \frac{10}{3} \]
Step 4: Final Answer:

The incenter is \( \left(0, \frac{10}{3}\right) \).