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(A) Let the point $P$ be $(x, y)$.The equation of the tangent at $P$ is $Y - y = y'(X - x)$.For the $x$-axis,set $Y = 0$,which gives $X = x - \frac{y}

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length of latus rectum $3$

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(A) Let the point $P$ be $(x, y)$.The equation of the tangent at $P$ is $Y - y = y'(X - x)$.For the $x$-axis,set $Y = 0$,which gives $X = x - \frac{y}{y'}$.Thus,the point $Q$ is $\left(x - \frac{y}{y'}, 0\right)$.The $y$-axis bisects the segment $PQ$,so the $x$-coordinate of the midpoint of $PQ$ must be $0$.$\frac{x + (x - \frac{y}{y'})}{2} = 0$ $\Rightarrow 2x - \frac{y}{y'} = 0$ $\Rightarrow y' = \frac{y}{2x}$.Separating variables,we get $\frac{dy}{y} = \frac{dx}{2x}$.Integrating both sides,$\ln(y) = \frac{1}{2} \ln(x) + C$,which simplifies to $y^2 = kx$.Since the curve passes through $(3, 3)$,$3^2 = k(3) \Rightarrow k = 3$.Thus,the curve is $y^2 = 3x$.Comparing with $y^2 = 4ax$,we have $4a = 3$,so the length of the latus rectum is $3$ and the focus is $\left(\frac{3}{4}, 0\right)$.

$A$ particle is moving in the $xy$-plane along a curve $C$ passing through the point $(3, 3)$. The tangent to the curve $C$ at the point $P$ meets the $x$-axis at $Q$. If the $y$-axis bisects the segment $PQ$,then $C$ is a parabola with

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