Two parabolas have the same focus, namely the point $(3,-28).$ Their directrices are the $x$-axis and the $y$-axis, respectively

Question

Norma-Jean

21 day ago

11

Two parabolas have the same focus, namely the point $(3,-28).$ Their directrices are the $x$-axis and the $y$-axis, respectively

. Compute the slope of their common chord.

Mathematics

1 answer:

babunello [8.4K] · Answer 1 · 2026-03-06T15:51:13+03:00

Answer:

The slope for the common chord is: $-1$

Step-by-step explanation:

The focus is located at (a,b) = (3,-28)

For a parabola that has its directrix along the x-axis, the equation will be

$\left(x-a\right)^{2}+\left(y-b\right)^{2}=y^{2}$

$\left(x-a\right)^{2}+\left(y-b\right)^{2}-y^{2}=0$

For a parabola with a directrix situated on the y-axis, the equation will be

$\left(x-a\right)^{2}+\left(y-b\right)^{2}=x^{2}$

$\left(x-a\right)^{2}+\left(y-b\right)^{2}-x^{2}=0$

The common chord is the line that connects the points where the two parabolas intersect. To find the intersection, we can set the equations of the two parabolas equal!

In essence, the values of the two parabolas at the intersection point will coincide.

$\left(x-a\right)^{2}+\left(y-b\right)^{2}-y^{2}=\left(x-a\right)^{2}+\left(y-b\right)^{2}-x^{2}$

$\left(x-3\right)^{2}+\left(y+28\right)^{2}-y^{2}=\left(x-3\right)^{2}+\left(y+28\right)^{2}-x^{2}$

Now we can simplify the equation. (The terms (x-3) and (y+28) will cancel each other out when subtracting the corresponding terms)

$-y^{2}=-x^{2}$

$y=\sqrt{x^{2}}$

$y=\pm x$

Thus, we arrive at the equation for the common chord, but we must determine whether it is

$y=+ x$ or $y=-x$ .

This can be discerned by observing that the focus is positioned in the 4th quadrant of the xy-plane! The equation $y=-x$ generates a line present in both the 2nd and 4th quadrants.

Therefore, the slope of the common chord corresponds to the slope of the line $y=-x$

which is: $-1$