Question

The equation of chord of the ellipse \( \frac{x^2}{25} + \frac{y^2}{16} = 1 \) with (3, 1) as mid-point is

• A. \( 48x + 25y - 169 = 0 \)
• B. \( 25x + 5y - 125 = 0 \)
• C. \( 65x + 2y - 12 = 0 \)
• D. \( 45x + 4y - 135 = 0 \)

Hint:

For finding the equation of a chord with a given midpoint in an ellipse, use the relation \( T = S_1 \) where \( T \) is the chord equation and \( S_1 \) is the ellipse equation at the midpoint.

Answer 1

The Correct Option is A

The general equation of the ellipse is: \[ \frac{x^2}{25} + \frac{y^2}{16} = 1 \] The equation of a chord with midpoint \( (x_1, y_1) \) can be written as: \[ T = S_1 \] where \( T \) is the equation of the chord and \( S_1 \) is the value of the ellipse equation at the midpoint. For the given ellipse, the midpoint is \( (3, 1) \), so we substitute \( x_1 = 3 \) and \( y_1 = 1 \) into the equation of the ellipse: \[ S_1 = \frac{3^2}{25} + \frac{1^2}{16} = \frac{9}{25} + \frac{1}{16} \] Now, find the least common denominator: \[ S_1 = \frac{144}{400} + \frac{25}{400} = \frac{169}{400} \] Thus, the equation of the chord is: \[ T = \frac{x^2}{25} + \frac{y^2}{16} - \frac{169}{400} = 0 \] Multiply through by 400 to eliminate the denominator: \[ 16x^2 + 25y^2 - 169 = 0 \] Thus, the equation of the chord is: \[ 48x + 25y - 169 = 0 \]