Two parabolas have the same focus (4, 3) and | vedclass

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(A) The definition of a parabola is the locus of points equidistant from the focus and the directrix. Let $P(x, y)$ be a point on the parabola.For the

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$192$

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(A) The definition of a parabola is the locus of points equidistant from the focus and the directrix. Let $P(x, y)$ be a point on the parabola.For the first parabola with focus $(4, 3)$ and directrix $y = 0$ ($x$-axis),the equation is $(x - 4)^2 + (y - 3)^2 = y^2$.Expanding this: $(x - 4)^2 + y^2 - 6y + 9 = y^2$,which simplifies to $(x - 4)^2 - 6y + 9 = 0$ or $6y = (x - 4)^2 + 9$.For the second parabola with focus $(4, 3)$ and directrix $x = 0$ ($y$-axis),the equation is $(x - 4)^2 + (y - 3)^2 = x^2$.Expanding this: $x^2 - 8x + 16 + (y - 3)^2 = x^2$,which simplifies to $(y - 3)^2 - 8x + 16 = 0$ or $8x = (y - 3)^2 + 16$.At the intersection points $A$ and $B$,both equations hold. Subtracting the two equations: $(x - 4)^2 - (y - 3)^2 = y^2 - x^2$,which simplifies to $x^2 - 8x + 16 - y^2 + 6y - 9 = y^2 - x^2$,or $2x^2 - 2y^2 - 8x + 6y + 7 = 0$. This does not immediately yield $x=y$. Let's re-evaluate: the intersection points satisfy $(x-4)^2 + (y-3)^2 = x^2$ and $(x-4)^2 + (y-3)^2 = y^2$,implying $x^2 = y^2$,so $y = x$ or $y = -x$. Since the focus $(4, 3)$ is in the first quadrant,the parabolas intersect where $y = x$.Substituting $y = x$ into $(x - 4)^2 + (x - 3)^2 = x^2$:$x^2 - 8x + 16 + x^2 - 6x + 9 = x^2$$x^2 - 14x + 25 = 0$.Let the roots be $x_1$ and $x_2$. Then $x_1 + x_2 = 14$ and $x_1 x_2 = 25$.The points are $A(x_1, x_1)$ and $B(x_2, x_2)$.$(AB)^2 = (x_1 - x_2)^2 + (x_1 - x_2)^2 = 2(x_1 - x_2)^2$.$(x_1 - x_2)^2 = (x_1 + x_2)^2 - 4x_1 x_2 = 14^2 - 4(25) = 196 - 100 = 96$.$(AB)^2 = 2 	imes 96 = 192$.

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

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