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What are the conic sections of the following equations $4 x^{2} + 4 y^{2} - 60 = 0$ $4x ^2 + 4y ^2 - 60=0$ ?

1 Answer

Jun 5, 2018

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Explanation:

$4 x^{2} + 4 y^{2} - 60 = 0$ $4x^2+4y^2-60=0$ dividing by 4

$x^{2} + y^{2} - \frac{60}{4} = x^{2} + y^{2} - 15 = 0$ $x^2+y^2-60/4=x^2+y^2-15=0$

This is a circle centered in $(0, 0)$ $(0,0)$ and radius $\sqrt{15}$ $sqrt15$

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