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I am given Focus (0,0) Directrix Y=3 eccentricity = 2

My rough sketch -

enter image description here

D2 = 2D1

$\sqrt{x^{2}+y^{2}}=2\ \sqrt{\left(y-3\right)^{2}}$

and simplifies to

$x^{2}-3y^{2}+24y=36$

The correct answer (from text book) shows as $x^{2}-3y^{2}+16y=16$

By graphing both the original equation and the result after manipulation, I see the result is consistent. Meaning my original set up was wrong. I am helping a precalc student with a series of these, and all others (6 with various combination of X or Y directrix) worked to match the book answer.

JTP - Apologise to Monica
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  • You draw a parabola, not an hyperbola. – Pspl Dec 04 '19 at 15:13
  • The sketch was one branch, of course. The second branch can only be drawn once the equation is known. A parabola has eccentricity 1, therefore D1=D2. – JTP - Apologise to Monica Dec 04 '19 at 15:14
  • What I mean was, draw the entire hyperbola (you don't have to place values on the branch you know nothing about), and maybe you'll see what you're doing wrong :) – Pspl Dec 04 '19 at 16:06
  • I appreciate the response, but, no, my image was edited, I only posted the bottom here. The relationship is just this, no? D2= E * D1 ? Sorry to be dense here, what am I missing? The mirror image above this gave me no idea, as I don't know the other directrix location. – JTP - Apologise to Monica Dec 04 '19 at 16:41
  • Please, take my advice and then calculate the directrix of your book answer (of the bottom branch), and you will see that you're right (and your book is wrong)... :) – Pspl Dec 04 '19 at 16:52
  • Disclosure - I work in a high school as an in-house math tutor (after retiring from a different career 7 years ago). I’m pretty good, but humble enough to know we all make mistakes. I see a book typo every now and then, but never assume I’m right and the book, wrong. (Yes, I’ll take your advice and calculate the book-answer e value to show the student I’m working with.) much thanks. – JTP - Apologise to Monica Dec 04 '19 at 17:04
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    I completely understand your frustration. I mean, come on, I double check, no... I triple check the results (and then I triple check again, and again...) when I start to assume the book is wrong and I'm right! But the truth is: sometimes it happens, eh eh! Good luck! :) – Pspl Dec 05 '19 at 12:35

1 Answers1

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Let me confirm your result by writing $$x^2-3y^2+24y-36~~~~(1)$$ in standard form $$\frac{(y-4)^2}{4}-\frac{x^2}{12}=1,$$ which is vertical hyperbola. its excentricity is $$e=\sqrt{\frac{4+12}{4}}=2$$ its transverse axis is $x=0$ and the conjugate axix is $y=4.$ The directrix is $ y-4=\pm 2/2 \implies y=5,3$ its focui are give by $y-4=\pm 2.2,$ $x=0$ so the foci are given by $x=0$ where $y=0,8$. The foci are $(0,0),(0,8)$.

So one directrix of the hyperbola is $y=3$ and its ecentricity is 2 one focus is $(0,0).$ Finally, your answer (1) is correct. See the Fig.

enter image description here

Z Ahmed
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