What kind of conic is defined by the equation $2 x^{2} + 4 y^{2} - 4 x + 12 y = 0$ $2x^2+4y^2-4x+12y=0$ ?

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What kind of conic is defined by the equation $2 x^{2} + 4 y^{2} - 4 x + 12 y = 0$ $2x^2+4y^2-4x+12y=0$ ?

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Answer 1 · 2017-10-24T12:17:47

the conic is an ellipse with centre $(1, - \frac{3}{2})$ $(1,-3/2)$ and eccentricity $\frac{1}{\sqrt{2}}$ $1/sqrt2$

Explanation:

the given equation is $2 x^{2} + 4 y^{2} - 4 x + 12 y = 0$ $2x^2+4y^2-4x+12y=0$
we are basically trying to convert the equation in form of perfect squares.
on doing so the equation can be written as $2 (x^{2} - 2 x + 1) + 4 (y^{2} + 3 y + {(\frac{3}{2})}^{2}) = 2 + 9$ $2(x^2-2x+1)+4(y^2+3y+(3/2)^2)=2+9$
$\Rightarrow 2 {(x - 1)}^{2} + 4 {(y + \frac{3}{2})}^{2} = 11$ $rArr2(x-1)^2+4(y+3/2)^2=11$
$\Rightarrow \frac{2 {(x - 1)}^{2}}{11} + \frac{4 {(y + \frac{3}{2})}^{2}}{11} = 1$ $rArr(2(x-1)^2)/11+(4(y+3/2)^2)/11=1$
$\Rightarrow \frac{{(x - 1)}^{2}}{\frac{11}{2}} + \frac{{(y + \frac{3}{2})}^{2}}{\frac{11}{4}} = 1$ $rArr(x-1)^2/(11/2)+(y+3/2)^2/(11/4)=1$
this is of the form $\frac{{(x - h)}^{2}}{a^{2}} + \frac{{(y - k)}^{2}}{b^{2}} = 1$ $((x-h)^2)/a^2+((y-k)^2)/b^2=1$
which is the general form of an elipse
hence centre is $(h, k) = (1, - \frac{3}{2})$ $(h,k)=(1,-3/2)$ and eccentricity $e = \sqrt{\frac{a^{2} - b^{2}}{a^{2}}} = \frac{1}{\sqrt{2}}$ $e=sqrt((a^2-b^2)/a^2)=1/sqrt2$

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What kind of conic is defined by the equation $2 x^{2} + 4 y^{2} - 4 x + 12 y = 0$ $2x^2+4y^2-4x+12y=0$ ?

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What kind of conic is defined by the equation $2 x^{2} + 4 y^{2} - 4 x + 12 y = 0$ $2x^2+4y^2-4x+12y=0$ ?