Question

One of the vertices of the major axis of an ellipse is (1, 1) and one of the vertices of its minor axis is (-2, -1). If the centre of the ellipse is (-2, 1), then the equation of the ellipse is

• A. \(\frac{(x+2)^2}{9}+\frac{(y-1)^2}{4}=1\)
• B. \(\frac{(x+2)^2}{16}+\frac{(y-1)^2}{4}=1\)
• C. \(\frac{(x-2)^2}{9}+\frac{(y+1)^2}{4}=1\)
• D. \(\frac{(x-2)^2}{16}+\frac{(y+1)^2}{4}=1\)
• E. \(\frac{(x+2)^2}{9}+\frac{(y-1)^2}{2}=1\)

Answer 1

The Correct Option is A

Step 1: Identify the standard form of the ellipse 
The standard form of the ellipse is: \[ \frac{(x - h)^2}{a^2} + \frac{(y - k)^2}{b^2} = 1 \] where \( (h, k) \) is the center, \( a \) is the semi-major axis, and \( b \) is the semi-minor axis. 
Given center: \( (-2,1) \), so the equation becomes: \[ \frac{(x + 2)^2}{a^2} + \frac{(y - 1)^2}{b^2} = 1 \]

Step 2: Find the lengths of the semi-major and semi-minor axes  
One vertex of the major axis is \( (1,1) \), so the distance from the center is: \[ |1 - (-2)| = |1 + 2| = 3 \] This is the semi-major axis, so \( a = 3 \) and \( a^2 = 9 \). 
One vertex of the minor axis is \( (-2,-1) \), so the distance from the center is: \[ |1 - (-1)| = |1 + 1| = 2 \] This is the semi-minor axis, so \( b = 2 \) and \( b^2 = 4 \).

Step 3: Write the final equation 
Substituting \( a^2 = 9 \) and \( b^2 = 4 \) into the equation: \[ \frac{(x + 2)^2}{9} + \frac{(y - 1)^2}{4} = 1 \]

Final Answer: The equation of the ellipse is \(\frac{(x+2)^2}{9}+\frac{(y-1)^2}{4}=1\).