Question

The abscissa of a point on the ellipse $\frac{x^2}{4}+\frac{y^2}{3}=1$ at a distance of $\frac{5}{4}$ unit from focus is

• A. $\pm \frac{3}{2}$
• B. $\pm \frac{1}{2}$
• C. $\pm \frac{1}{3}$
• D. $\pm \frac{1}{4}$

Answer 1

The correct option is A $\pm \frac{3}{2}$

For given ellipse $\frac{x^2}{4}+\frac{y^2}{3}=1$
$e=\sqrt{1-\frac{b^2}{a^2}}$
$\Rightarrow e=\sqrt{1-\frac{3}{4}}=\frac{1}{2}$
For focal distance
$\because a\pm e{x}_1=\frac{5}{4}$
$\Rightarrow 2\pm \frac{1}{2}{x}_1=\frac{5}{4}$
$\Rightarrow x_1=\pm \frac{3}{2}$