Question

The centre of the ellipse lies on the lines \(2x+3y=5\) and \(x+3y=4\). If the eccentricity of the ellipse is \(\frac23\), length of its major axis is 4 and its minor axis is parallel to Y-axis, then the equation of the ellipse is:

• A. \(5(x-1)^2+9(y-1)^2=20\)
• B. \(5(x+1)^2+9(y+1)^2=20\)
• C. \(9(x-1)^2+5(y-1)^2=20\)
• D. \(9(x+1)^2+5(y+1)^2=20\)

Hint:

If the minor axis is parallel to Y-axis, then the major axis must be parallel to X-axis.

Answer 1

The Correct Option is A

Concept: The centre of the ellipse is obtained by solving the two given lines. Since the minor axis is parallel to Y-axis, the major axis is parallel to X-axis.

Step 1: Find the centre.
Solve \[ 2x+3y=5 \] and \[ x+3y=4. \] Subtracting, \[ x=1. \] Substituting, \[ 1+3y=4 \] \[ y=1. \] Thus centre \[ (1,1). \]

Step 2: Find semi-major and semi-minor axes.
Major axis length \[ 2a=4. \] Hence \[ a=2. \] Given \[ e=\frac23. \] \[ e^2=1-\frac{b^2}{a^2}. \] \[ \frac49 = 1-\frac{b^2}{4}. \] \[ \frac{b^2}{4} = \frac59. \] \[ b^2=\frac{20}{9}. \]

Step 3: Form equation.
\[ \frac{(x-1)^2}{4} + \frac{(y-1)^2}{20/9} = 1. \] Multiplying by 20, \[ 5(x-1)^2+9(y-1)^2=20. \] Hence \[ \boxed{5(x-1)^2+9(y-1)^2=20}. \]