Two parabolas have the same focus $(4, 3)$ and their directrices are the $x$-axis and the $y$-axis, respectively. If these parabolas intersect at the points $A$ and $B$, then $(AB)^2$ is equal to
Step-by-Step Solution
Key Concept: Write the two parabola equations via $PS=PM$; subtract them to get $x^2=y^2$, so $x=y$ (only real branch); then reduce to a single quadratic in $x$ and use Vieta's formulas to find $(AB)^2 = 2[(x_1+x_2)^2-4x_1x_2]$.
Parabola 1 (directrix $y=0$): $(x-4)^2+(y-3)^2=y^2$. Parabola 2 (directrix $x=0$): $(x-4)^2+(y-3)^2=x^2$. Subtracting: $x^2=y^2 \Rightarrow x=y$ (the branch $x=-y$ gives no real solutions). Substituting $y=x$ into Parabola 2: $x^2-14x+25=0$, so $x_1+x_2=14$, $x_1 x_2=25$. Since $A=(x_1,x_1)$ and $B=(x_2,x_2)$: $$(AB)^2=2(x_1-x_2)^2=2\bigl[(x_1+x_2)^2-4x_1x_2\bigr]=2(196-100)=192.$$
Correct Answer: 3