Question

The equation of the ellipse whose vertices are (–4, 1), (6, 1) and one latus rectum is x – 4 = 0 is

• A. $\frac{(x-1{)}^2}{25}+\frac{(y-1{)}^2}{16}=1$
• B. $\frac{(x-1{)}^2}{16}+\frac{(y-1{)}^2}{25}=1$
• C. $\frac{(x+1{)}^2}{25}+\frac{(y+1{)}^2}{16}=1$
• D. $\frac{(x+1{)}^2}{16}+\frac{(y+1{)}^2}{25}=1$

Answer 1

The correct option is A $\frac{(x-1{)}^2}{25}+\frac{(y-1{)}^2}{16}=1$

A = (–4, 1), A’ = (6, 1) are the vertices of an ellipse.
Centre = C = midpoint of $A{A}^{\prime }=(\frac{-4+6}{2},\frac{1+1}{2})=(1,1)=(\alpha ,\beta )$
Since y-coordinates are equal in A, A’ the major axis is parallel to x-axis
Length of the major axis 2a = AA’
a = 5
Distance between centre and the focii = ae = 5
e = $\frac{1}{5}$
So the equation of the ellipse is $\frac{(x-1{)}^2}{25}+\frac{(y-1{)}^2}{16}=1$.