Question

An ellipse has its centre at (1,-1) and semi-major axis=8 and it passes through the point (1,3). The equation of the ellipse is

• A. $\frac{(x+1{)}^2}{64}+\frac{(y+1{)}^2}{16}=1$
  
• B. $\frac{(x-1{)}^2}{64}+\frac{(y+1{)}^2}{16}=1$
  
• C. $\frac{(x-1{)}^2}{16}+\frac{(y+1{)}^2}{64}=1$
  
• D. $\frac{(x-1{)}^2}{64}+\frac{(y+1{)}^2}{16}=1$

Answer 1

The correct option is B $\frac{(x-1{)}^2}{64}+\frac{(y+1{)}^2}{16}=1$


According to the question, the centre is at (1,-1)
a=8 (Given)
$\Rightarrow \frac{(x-1{)}^2}{a^2}+\frac{(y+1{)}^2}{b^2}=1$ ...(1)
It passes through point (1,3)
i.e., x=1 and y=3
Putting these values in eq. (1), we get:
$\frac{(1-1{)}^2}{a^2}+\frac{(3+1{)}^2}{b^2}=1$
$\Rightarrow \frac{16}{b^2}=1$
$\Rightarrow b^2=16orb=4$
substituting the values of a and b in eq. (1), we get:
$\frac{(x-1{)}^2}{64}+\frac{(y+1{)}^2}{16}=1$