Question

The parabola $y^2=px$ passes through the point of intersection of the lines $\frac{x}{3}+\frac{y}{2}=1$ and $\frac{x}{2}+\frac{y}{3}=1$. Its focus is

• A. $\left(\frac{3}{10},0\right)$
• B. $\left(\frac{3}{5},0\right)$
• C. $\left(\frac{3}{7},0\right)$
• D. $\left(\frac{6}{7},0\right)$

Answer 1

The correct option is A $\left(\frac{3}{10},0\right)$

Given parabola is $y^2=px$ ..... $\left(i\right)$

$\frac{x}{3}+\frac{y}{2}=1$ ..... $\left(ii\right)$
$\frac{x}{2}+\frac{y}{3}=1$ ..... $\left(iii\right)$

Solving above equations, we get
$x=\frac{6}{5},y=\frac{6}{5}$

From $\left(i\right)$, we get
${\left(\frac{6}{5}\right)}^2=p\times \left(\frac{6}{5}\right)$
$p=\frac{6}{5}$

Again, from $\left(i\right)$
$4a=\frac{6}{5}$
$a=\frac{3}{10}$
Focus $=(a,0)$ $=(\frac{3}{10},0)$