The equation is x²-y²+6x+2y=10. What is the center,vertices,foci length of Latus Rectum, asymptotes and the graph of hyperbola?

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Answer 1 · 2018-07-22T00:22:39

See below

Let us start by putting this equation in its standard form:
$x²-y²+6x+2y=10$

$x^2+6x-y^2+2y=10$

$(x+3)^2-(y-1)^2=10+9-1$

$(x+3)^2/(18)-(y-1)^2/(18)=1$

$vertex: (-3, 1)$

$a= sqrt18=3sqrt2$
$b= sqrt18= 3sqrt2$

$c= sqrt(a^2+b^2)$

$c= sqrt36= 6$

To get vertices, add and subtract $a$ to the $x$ coordinates of the vertex:
$Vertices: (-3+3sqrt2, 1) and (-3-3sqrt2, 1)$

To get focii, add and subtract the $c$ to the $x$ coordinates of the vertex:
$Focii: (3, 1) and (-9, 1)$

Length of the latus rectum:
$L=2b^2/a= (2*18)/(3sqrt2)=6sqrt2$
Now graph these points using estimations

For the asymptotes, simply go $b/a$ in both directions from the center:
$y=x+4$
$y=-x-2$