JEE Main · 2022 · 24 Jun 2022 · Shift 2 · Parabola

Let $P(x,y)$ be a point on the curve. The tangent at P has equation:

$Y-y=y^{\prime }(X-x)$

At the x-axis ($Y=0$): $X=x-\frac{y}{y^{\prime }}$, so $Q\left(x-\frac{y}{y^{\prime }},0\right)$

Since the y-axis bisects PQ, the midpoint has x-coordinate 0:

$\frac{x+(x-y/y^{\prime })}{2}=0\Rightarrow 2x=\frac{y}{y^{\prime }}$

$y^{\prime }=\frac{y}{2x}$

Integrating: $2\ln y=\ln x+\ln k$, giving $y^2=kx$

Since the curve passes through $(3,3)$: $9=3k$, so $k=3$

The curve is $y^2=3x$, which has latus rectum length 3 and focus $\left(\frac{3}{4},0\right)$.