1. Set up the condition using a general point on the curve

Let $P(x,y)$ be any point on the curve $C$, and let the tangent at $P$ meet the $x$-axis at $Q$.

Given that the curve passes through $(3,3)$.

If the slope of the tangent at $P$ is $\dfrac{dy}{dx}=m$, then the tangent at $P(x,y)$ is

$Y-y=m(X-x).$

To find $Q$, put $Y=0$ (since $Q$ lies on the $x$-axis):

$-y=m(X-x)$ $X=x-\frac{y}{m}=x-\frac{y}{dy/dx}.$

So,

$Q\left(x-\frac{y}{dy/dx},\,0\right).$

2. Use the midpoint condition

The $y$-axis bisects the segment $PQ$. Since the $y$-axis is $x=0$, the midpoint of $P(x,y)$ and $Q\left(x-\frac{y}{y'},0\right)$ must have $x$-coordinate $0$.

Thus,

$\frac{x+\left(x-\frac{y}{y'}\right)}{2}=0.$

So,

$2x-\frac{y}{y'}=0$ $\frac{y}{y'}=2x$ $y'=\frac{y}{2x}.$

Hence the differential equation of the curve is

$\frac{dy}{dx}=\frac{y}{2x}.$

3. Solve the differential equation

Separate variables:

$\frac{dy}{y}=\frac{dx}{2x}.$

Integrating,

$\ln y=\frac{1}{2}\ln x + C$ $y=C\sqrt{x}.$

Squaring,

$y^2=C^2x.$

So the curve is a parabola of the form

$y^2=4ax.$

4. Use the point $(3,3)$

Since $(3,3)$ lies on the curve,

$3^2=C^2\cdot 3$ $9=3C^2$ $C^2=3.$

Therefore,

$y^2=3x.$

Comparing with $y^2=4ax$,

$4a=3 \quad \Rightarrow \quad a=\frac{3}{4}.$

5. Find latus rectum and focus

For the parabola $y^2=4ax$:

• Focus is $(a,0)$
• Length of latus rectum is $4a$

Here $a=\dfrac{3}{4}$, so

• Focus $=\left(\dfrac{3}{4},0\right)$
• Length of latus rectum $=4a=3$

6. Check the options

• A: length of latus rectum $3$ ✅
• B: length of latus rectum $6$ ❌
• C: focus $\left(\dfrac{4}{3},0\right)$ ❌
• D: focus $\left(0,\dfrac{3}{4}\right)$ ❌

Therefore, the correct option is A.