Question

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.

Answer 1

Answer:

the slope of the common cord is: $-1$

Step-by-step explanation:

Given the focus (a,b) = (3,-28)

• for a parabola with directrix at 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 directrix at 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 between two points where the two parabolas intersect. For intersection, we can equate the two parabolas!

In other words, at the point of intersection of these two parabolas the values of the two parabolas will be the same.

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

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

we can now simplify the equation. (we can see that (x-3) and (y+28) both cancel out by -(x-3) and -(y+28))

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

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

$y=\pm x$

this the equation of the common cord. but we need to select whether its

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

This can be found by realizing that the focus lies on the 4th quadrant of the xy-plane! And the equation $y=-x$ also generates a line that exists in the 2nd and 4th quadrant.

Hence the slope of the common cord is the slope of the line $y=-x$

that is : $-1$