3

For a National Board Exam Review:

Compute the eccentricity of a given curve $9x^2 + 4y^2 - 24y + 144 = 0$

Answer is $0.75$

I try:

$$9x^2 + 4y^2 - 24y + 144 = 0$$ $$9x^2 + 4(y^2 - 6y + 9) = -144 + 36$$ $$9x^2 + 4(y-3)^2 = -108$$ $$\frac{-x^2}{12} - \frac{4(y-3)^2}{27} = 1$$

For an ellipse:

$$ a^2 - b^2 = c^2 $$ $$ 12 - 27 = 15 $$

$$ e = \frac{c}{a} = \frac{ \sqrt{15} }{ \sqrt{12} } = \frac{\sqrt{5}}{2} $$

It's also odd because I have never encountered a conic which has both terms negative. What am I doing wrong?

Michael Hardy
  • 1
james
  • 1,017
  • On line two, you subtract off 144, but you still have a constant on the left side. – Michael Dyrud Aug 21 '15 at 00:38
  • @MichaelDyrud sorry typo. corrected – james Aug 21 '15 at 00:39
  • You also need to make sure you've factored out a 4 from the -24 on the $y$ term. – Michael Dyrud Aug 21 '15 at 00:39
  • 4
    Check it's not $-144$ in the question... I've just sketched it with "$-144$" and that looks much better... – tomi Aug 21 '15 at 00:46
  • 5
    Unless $x$ and $y$ can be complex numbers, there are no $x$ and $y$ that solve the equation $$\frac{x^2}{12}+\frac{(y-3)^2}{27}=-1$$ as both terms on the left hand side have to be bigger than or equal to $0$. – J126 Aug 21 '15 at 00:56
  • Per Joe Johnson 126's remark, either you've copied the problem wrong, or else the exam was in error. A different problem with your calculation is that the semi-major axis is the larger of the two values, i.e, it is $\sqrt{27}$, not $\sqrt{12}$. – Paul Sinclair Aug 21 '15 at 01:40

0 Answers0