Question

Two parabola have the focus $(3,-2)$. Their directrices are the $x$-axis, and $y$-axis respectively. Then the slope of their common chord is

• A. $-1$
• B. $-\frac{1}{2}$
• C. $+1$
• D. both $A$ and $B$

Answer 1

The correct option is D both $A$ and $B$

Let the Parabolas with axis as $X-$ and $Y-$ be $P_x$ and $P_y$.

$S(3,2)$

$=>P_x:{(x-3)}^2+({y+2)}^2=\frac{{\left(x\right)}^2}{1+0}$

$=>{(y+2)}^2=6(x-\frac{3}{2})$

$=>y^2+4y+13-6x=0......\left(i\right)$

$=>P_y:{(x-3)}^2+{(y+2)}^2=y^2$

$=>{(x-3)}^2=-4(y+1)$

$=>x^2-6x+4y+13=\mathrm{0........}\left(ii\right)$

Point of intersection of $\left(i\right)$ and $\left(ii\right)$ $=>x^2=y^2$

$=>y=\pm x$

$x=y$ and $y=-x$ are chords to $P_x,P_y$

So, slope chords is $-1$ and $1$.