Line $x-1=0$ is directrix of parabola $y^2-kx +8=0$ that intersects circle $x^2+y^2=4$ at two distinct real points

Question

The parabola intersects the circle $x^2+y^2=4$ at two distinct real point, then find k.

$$y^2=kx-8$$ $$y^2=4(\frac k4)(x-\frac 8k)$$

hence $$\frac k4=-1$$ $$k=-4$$ The right answer is +4. I now understand it has got something to do with parabola repointing either way, but that’s because I the answer. Is there any way to find it out purely using mathematics?

In other words, I got -4, but the correct answer is +4. What’s wrong?

Answer 1

Note that there are two parabolas that have $x-1=0$ as their directrix, i.e.

$$y^2=4x-8,\>\>\>\>\>\>\>y^2=-8x-8$$

Then, you need to examine their intersecting points with the circle $x^2+y^2=4$ to determine which one satisfies the requirement of two distinctive real points.

It turns out that the parabola $y^2=4x-8$ has only one intercept with the circle at $(2,0)$, while the other parabola $y^2=-8x-8$ has two intercepts, at $(-2\sqrt7+4, \pm 2\sqrt{4\sqrt7-10})$. Thus, the second parabola satisfies the condition, i.e. $k=-8$.

Answer 2

Question: If the line $~x - 1 = 0~$ is the directrix of the parabola $~y^2 - kx + 8 = 0~$, then what is the values of $k$ ?

Answer: The equation of the parabola can be written as $$y^2 - kx + 8 = 0\implies y^2 = kx -8 \implies y^2=k\left(x-\dfrac 8k\right)$$which is of the form$~Y^2=kX~$ where $~Y=y~~$and$~~X=x-\dfrac 8k~$.

In standard form it can be written as $$Y^2=kX\implies Y^2=4~\dfrac k4 X$$so that $~a = \dfrac k4~$ and the equation of the directrix is $~X + a = 0 ⇒ x-\left(\dfrac 8k - \dfrac k4\right) = 0~$.

Now given that $~x – 1 = 0~$ is the directrix

So $~\dfrac 8k-\dfrac k4=1\implies k=4~,-8~.$

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Question: The parabola $~y^2 - kx + 8 = 0~$ intersects the circle $~x^2+y^2=4~$ at two distinct real point, then find $~k~$.

Answer: If the parabola $~y^2 - kx + 8 = 0~$ intersects the circle $~x^2+y^2=4~$, then $$x^2+(kx-8)=4\implies x^2+kx-12=0\implies x=\dfrac{-k\pm \sqrt{k^2+48}}{2}$$For distinct, real solution of the quadratic equation $~ax^2 + bx + c = 0~$ (where $~a,~ b~$ and $~c~$ are real numbers, $~a ≠ 0~$), $~b^2 - 4ac > 0~$.

Here $~~k^2+48\gt 0~$ for all value of $~k~$.