If the line $x-1=0$ is the directrix of the parabola $y^2-kx+8=0$, then find the value of k

Question

The parabola is $$y^2=4\frac k4 (x-\frac 8k)$$

The directrix for this parabola is $$x-\frac 8k+\frac k4=0$$

Then $$\frac k4 -\frac 8k=1$$ $$k^2-4k-32=0$$ $$k=8,-4$$

But the answer given is 4. I checked my computation, but there doesn’t seem to be anything wrong. Maybe there is a problem with the signs ie.$y^2=\pm 4ax$

What is the actual problem.

What you did looks fine to me. – José Carlos Santos Mar 08 '20 at 17:03 — José Carlos Santos, Mar 08 '20 at 17:03
@JoséCarlosSantos no it’s a computation error – Aditya Mar 08 '20 at 17:49 — Aditya, Mar 08 '20 at 17:49
LHS of third equation has the wrong sign. – Intelligenti pauca Mar 08 '20 at 17:50 — Intelligenti pauca, Mar 08 '20 at 17:50

score 1 · Accepted Answer · edited Mar 08 '20 at 17:47

$$y^2=kx-8 \implies y^2=k(x-8/k)$$ Comparing it with $Y*2=4A X$, we get the length of larectum $4A=k$. The equation of the directrix is $$X=-A \implies (x-8/k)=-k/4 \implies x-8/k+k/4$$ comparing it with $x=1$, we get $$8/k-k/4=1 \implies k^2+4k-32=0 \implies k=4, -8$$. In both the cases $x=1$ is the directrix. For $k=4$ (blue) parabola is right facing and for $k=-8$ it is facing left (red). See the Fig. below.

If the line $x-1=0$ is the directrix of the parabola $y^2-kx+8=0$, then find the value of k

1 Answers1