Question

Let the ellipse $ 3x^2 + py^2 = 4 $ pass through the centre $ C $ of the circle $ x^2 + y^2 - 2x - 4y - 11 = 0 $ of radius $ r $. Let $ f_1, f_2 $ be the focal distances of the point $ C $ on the ellipse. Then $ 6f_1 f_2 - r $ is equal to

• A. \(70\)
• B. \(68\)
• C. \(78\)
• D. \(74\)

Hint:

For ellipses, remember: - The standard form \( \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \). - The focal distance \( c \) is given by \( c^2 = |a^2 - b^2| \). - The sum of focal distances for any point on the ellipse is \( 2b \) (if \( b > a \)).

Answer 1

The Correct Option is A

Step 1: Find the center \( C \) of the circle. The given circle equation is: \[ x^2 + y^2 - 2x - 4y - 11 = 0. \] Rewrite it in standard form by completing the square: \[ (x^2 - 2x) + (y^2 - 4y) = 11, \] \[ (x^2 - 2x + 1) + (y^2 - 4y + 4) = 11 + 1 + 4, \] \[ (x - 1)^2 + (y - 2)^2 = 16. \] Thus, the center \( C \) is at \( (1, 2) \), and the radius \( r = 4 \).
Step 2: Substitute \( C \) into the ellipse equation to find \( p \). The ellipse equation is: \[ 3x^2 + py^2 = 4. \] Substitute \( C = (1, 2) \): \[ 3(1)^2 + p(2)^2 = 4 \quad \Rightarrow \quad 3 + 4p = 4 \quad \Rightarrow \quad p = \frac{1}{4}. \] So, the ellipse becomes: \[ 3x^2 + \frac{1}{4}y^2 = 4 \quad \Rightarrow \quad \frac{x^2}{\frac{4}{3}} + \frac{y^2}{16} = 1. \]
Step 3: Identify the semi-major and semi-minor axes. The standard form of the ellipse is: \[ \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1, \] where \( a^2 = \frac{4}{3} \) and \( b^2 = 16 \).
Since \( b>a \), the major axis is along the \( y \)-axis.
The focal distance \( c \) is given by: \[ c^2 = b^2 - a^2 = 16 - \frac{4}{3} = \frac{44}{3} \quad \Rightarrow \quad c = \frac{2\sqrt{33}}{3}. \]
Step 4: Calculate the focal distances \( f_1 \) and \( f_2 \) for point \( C \). For an ellipse \( \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \) with \( b>a \), the focal distances of a point \( (x_0, y_0) \) are: \[ f_1 = b + \frac{c}{b} y_0, \quad f_2 = b - \frac{c}{b} y_0. \] Substitute \( C = (1, 2) \), \( b = 4 \), and \( c = \frac{2\sqrt{33}}{3} \): \[ f_1 = 4 + \frac{\frac{2\sqrt{33}}{3}}{4} \cdot 2 = 4 + \frac{\sqrt{33}}{3}, \] \[ f_2 = 4 - \frac{\frac{2\sqrt{33}}{3}}{4} \cdot 2 = 4 - \frac{\sqrt{33}}{3}. \] Now, compute \( f_1 f_2 \): \[ f_1 f_2 = \left(4 + \frac{\sqrt{33}}{3}\right)\left(4 - \frac{\sqrt{33}}{3}\right) = 16 - \left(\frac{\sqrt{33}}{3}\right)^2 = 16 - \frac{33}{9} = 16 - \frac{11}{3} = \frac{37}{3}. \]
Step 5: Compute \( 6f_1 f_2 - r \). \[ 6f_1 f_2 - r = 6 \cdot \frac{37}{3} - 4 = 74 - 4 = 70. \]