Equation of hyperbola confocal with ellipse having same principal axes

  • #1
Aurelius120
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Homework Statement
If the ellipse $$4x^2+9y^2+12x+12y+5=0$$ is confocal with a hyperbola having same principal axes then:
Relevant Equations
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The equation of ellipse reduces to :
$$(2x+3)^2+(3y+2)^2=8$$
$$\frac{(x+3/2)^2}{8/4}+\frac{(y+2/3)^2}{8/9}=1$$

Center of ellipse =##\left(\frac{-3}{2},\frac{-2}{3}\right)##

##b^2=a^2(1-e^2)=8/9## and ##a^2=8/4##
Therefore ##e=\frac{\sqrt{5}}{3}##

Distance between foci=##\frac{2\sqrt{10}}{3}##
Length of latus rectum of ellipse is 2 times the value of ##y## obtained on putting ##x=ae## and is equal to ##\frac{8\sqrt{2}}{9}##
Now comes the difficult part :
Finding the equation of hyperbola. I simply can't seem to do it.

For confocal hyperbola,
$$\frac{x^2}{a^2}-\frac{y^2}{b^2}=1\text{ and } ae=\frac{\sqrt{10}}{3}$$.
If principal axes have same length they become the same curve so I think principal axes are unequal in length but coincident.

I don't know what to do next ? I think there's insufficient information to get the equation of hyperbola.
 
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  • #2
Also just to clarify more than one options are correct so all need to be checked
 

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