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We have a point $A(6,0)$ and a line $k:y=2$. Show that the equation of the parabola with a locus $A$ and a directrix $k$ has the formula: $\dfrac{1}{4}x^2-3x+8$.

I had a test on analytic geometry today and this was one of the questions. I was 100% sure that the question was wrong, however, all my classmates told me that they had solved it correctly (apart from the fact that it should have been $-\dfrac{1}{4}$ instead of $\dfrac{1}{4}$, something which I noticed too, disregard this when answering my question). So my question is what I did incorrectly:

We use the formula of a parabola, $y-b=\dfrac{1}{4c}(x-a)^2$. The distance between the directrix and the locus is 2, so $c=1$. We get $y-b=\dfrac{1}{4}(x-a)^2$. We see that the top of the parabola is $T(6,1)$, so we get $y-1=\dfrac{1}{4c}(x-6)^2$. This can also be written as: $y=\dfrac{1}{4}x^2 -12x + 37$. And here we go, a totally different answer. Can anyone point out my mistakes?

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I suppose you meant "focus" instead "locus"...didn't you? –  DonAntonio Dec 4 '12 at 19:58
    
Indeed I did (at least I thought it was named focus). –  JohnPhteven Dec 4 '12 at 19:59
    
@DavidMitra Well, that doesn't help me, sir. –  JohnPhteven Dec 4 '12 at 20:03

2 Answers 2

up vote 1 down vote accepted

If the directrix is the line $y=2$:

Your value of $c$ should be negative, since the focus lies below the directrix. So, your initial equation should be $$\tag{1} y-1={-1\over 4}(x-6)^2.$$ Apart from $c=-1$, and not $c=1$, what you had there is correct.

But, it seems you made some algebraic errors when writing your initial equation in standard form. To write equation $(1)$ in standard form, you could first multiply both sides by $-4$ to obtain $$ -4y+4=(x-6)^2. $$ Now expand the right hand side: $$ -4y+4=x^2-12x+36. $$ So, subtracting $4$ from both sides, we obtain $$-4y=x^2-12 x +32.$$ Finally, divide through by $-4$ $$ y={-1\over4} x^2+3x-8. $$

But then, the proposed solution is incorrect; the right hand side should be multiplied by $-1$. (But the given answer would be correct if the directrix were the line $y=-2$.)

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I think the sign of c depend of the position of the focus compared to the directrix (question of orientation) because if you set it at -1 you will get the answer I compute (-1/4 for 1/4).

By the way, you forgot to divide by four the affine term.

Please excuse the deplorable english of a french (thought I hope it remains understandable).

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