Question

Let the line 2x + 3y – k = 0, k > 0, intersect the x-axis and y-axis at the points A and B, respectively. If the equation of the circle having the line segment AB as a diameter is x² + y² – 3x – 2y = 0 and the length of the latus rectum of the ellipse x² + 9y² = k² is m n , where m and n are coprime, then 2m + n is equal to

• A. 10
• B. 11
• C. 13
• D. 12

Answer 1

The Correct Option is B

Centre of the circle:

\[ \left( \frac{3}{2}, 1 \right) \]

Equation of diameter:

\[ 2\left( \frac{3}{2} \right) + 3(1) - k = 0 \implies k = 6 \]

Now, equation of ellipse becomes:

\[ x^2 + 9y^2 = 36 \]

\[ \frac{x^2}{6^2} + \frac{y^2}{2^2} = 1 \]

Length of latus rectum (LR):

\[ LR = \frac{2b^2}{a} = \frac{2 \cdot 2^2}{6} = \frac{8}{6} = \frac{4}{3} = \frac{m}{n} \]

Thus, \[ 2m + n = 2(4) + 3 = 11 \]

Answer 2

Step 1: Find intercepts of the line
The line is 2x + 3y – k = 0, k > 0.
x–intercept A: set y = 0 ⇒ 2x – k = 0 ⇒ x = k/2 ⇒ A = (k/2, 0).
y–intercept B: set x = 0 ⇒ 3y – k = 0 ⇒ y = k/3 ⇒ B = (0, k/3).

Step 2: Use the diameter condition of the given circle
The circle with diameter AB has equation x² + y² – 3x – 2y = 0.
Its center is the midpoint of A and B, i.e. M = (3/2, 1).
But the midpoint of A(k/2, 0) and B(0, k/3) is \[ \left(\frac{k/2 + 0}{2}, \frac{0 + k/3}{2}\right) = \left(\frac{k}{4}, \frac{k}{6}\right). \] Equate coordinates with (3/2, 1): k/4 = 3/2 ⇒ k = 6, and k/6 = 1 ⇒ k = 6 (consistent).

Step 3: Form the ellipse and its parameters
Given ellipse: x² + 9y² = k². With k = 6, it becomes x² + 9y² = 36 ⇒ \[ \frac{x^2}{36} + \frac{y^2}{4} = 1, \] so a² = 36, b² = 4 ⇒ a = 6, b = 2 (major axis along x).

Step 4: Length of the latus rectum
For ellipse \(\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1\) with a ≥ b, the length of the latus rectum is \[ \ell = \frac{2b^2}{a} = \frac{2 \cdot 4}{6} = \frac{8}{6} = \frac{4}{3}. \] Thus \(\ell = m/n\) with coprime m, n ⇒ m = 4, n = 3.

Step 5: Compute 2m + n
2m + n = 2·4 + 3 = 8 + 3 = 11.

Final answer

11