Question

What type of conic is this? What is the standard form of it? 2y^2 -x^2 + 2x +8y + 11=0

Answer 1

\[2y^2 -x^2 + 2x +8y + 11=0\]

Answer 2

Notice the \(-\) before the \(x^2\); this only happens if our function is of the form $$\frac{(y-k)^2}{a^2}-\frac{(x-h)^2}{b^2}=1$$... hmm, isn't that a hyperbola?

Converting to standard form merely requires completing the square in \(x\) and in \(y\).

Answer 3

I'll do \(x\) for you: $$2y^2-x^2+2x+8y+11=0\\2y^2+8y+11-(x^2-2x)=0\\2y^2+8y+11-(x^2-2x+1)+1=0\\2y^2+8y+12-(x-1)^2=0$$

Answer 4

Thanks, When you are left with
\[2y^2+8y+12−(x−1)^2=0\]
You can make it \[2(y^2+4y+6) −(x−1)^2=0\]

But then I'm not sure how you get it into (y-k)^2 /a^2 format..

Answer 5

The book tells me
a= 2
b= sqrt (2)
c= sqrt (6),
But how do i get that?

Answer 6

@toadytica305 leave it as \(2(y^2+4y)+12\) and complete the square of \(y\), so \(2(y^2+4y+4)+12-8=2(y+2)^2+4\). Our entire equation thus far is then $$2(y+2)^2-(x-1)^2+4=0\\2(y+2)^2-(x-1)^2=-4\\(y+2)^2-\frac{(x-1)^2}2=-2\\-\frac{(y+2)^2}2+\frac{(x-1)^2}4=1\\\frac{(x-1)^2}4-\frac{(y+2)^2}2=1$$ so then our vertex is \((h,k)=(1,-2)\) and our \(a=\sqrt4=2\), \(b=\sqrt2\). \(c\) can then be found by the Pythagorean theorem: $$c=\sqrt{a^2+b^2}=\sqrt{2+4}=\sqrt6$$