Question

In an ellipse, if centre is at origin, foci are $(\pm 3,0)$ and length of semi-major axis is 4 , then the equation of ellipse is

• A. $\frac{x^2}{16}+\frac{y^2}{7}=1$
• B. $\frac{x^2}{7}+\frac{y^2}{16}=1$
• C. $\frac{x^2}{16}+\frac{y^2}{49}=1$
• D. $\frac{x^2}{49}+\frac{y^2}{16}=1$

Answer 1

Answer: (A)

Since, foci $(\pm 3,0)$ lie on $X$-axis, as $y$-coordinate is zero.
Hence, equation of ellipse will be of the form
$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$...(i)
Given that, foci $(\pm c, 0)=(\pm 3,0)$
$c=3 \text { and } a=4$
$\because c^2 =a^2-b^2$
$\Rightarrow (3)^2 =(4)^2-b^2 $
$\Rightarrow 9 =16-b^2$
$\Rightarrow b^2 =16-9$
$\Rightarrow b^2 =7$
Put the values of $a^2=16$ and $b^2=7$ in Eq. (i), we get
$\frac{x^2}{16}+\frac{y^2}{7}=1$