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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?

Quanto
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Aditya
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2 Answers2

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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 $~k~$.

nmasanta
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  • So it doesn’t have anything to do with the circle? – Aditya Feb 03 '20 at 16:37
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    @Aditya: You'll have to explain why you've asked one question in the title, and another in the body. What is the connection you see between them? Or were you merely told there is a connection? – Brian Tung Feb 03 '20 at 16:40
  • Actually I think, when you copy the question then you combine the two separate questions.@Aditya – nmasanta Feb 03 '20 at 16:43
  • @BrianTung They are the same question. They are supposed to be connected, but could be irrelevant to confuse the person who solves it. – Aditya Feb 03 '20 at 16:45
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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. $k=-8$.

Quanto
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