How do you find all the critical points to graph $\frac{{(x + 2)}^{2}}{4} - \frac{{(y - 5)}^{2}}{25} = 1$ $(x + 2)^2/4 - (y - 5)^2/25 = 1$ including vertices, foci and asymptotes?

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Answer 1 · 2018-07-30T07:46:58

Please see the explanation below

The equation of a hyperbola with a horizontal transverse axis is

$\frac{{(x - h)}^{2}}{a^{2}} - \frac{{(y - k)}^{2}}{b^{2}} = 1$ $(x-h)^2/a^2-(y-k)^2/b^2=1$

Our equation is

$\frac{{(x + 2)}^{2}}{4} - \frac{{(y - 5)}^{2}}{25} = 1$ $(x+2)^2/4-(y-5)^2/25=1$

$a = 2$ $a=2$

$b = 5$ $b=5$

$h = - 2$ $h=-2$

$k = 5$ $k=5$

$c = \sqrt{b^{2} + c^{2}} = \sqrt{25 + 4} = \sqrt{29}$ $c=sqrt(b^2+c^2)=sqrt(25+4)=sqrt29$

The center is $C = (h, k) = (- 2, 5)$ $C=(h,k)=(-2,5)$

The vertices are $A = (h + a, k) = (0, 5)$ $A=(h+a,k)=(0,5)$

and

$C'=(h-a,k)=(-4,5)$

The foci are $F=(h+c,k)=(-2+sqrt29, 5)$

and

$F'=(-2-sqrt29, 5)$

$(x+2)^2/4-(y-5)^2/25=0$

$(x+2)^2/4=(y-5)^2/25$

$(y-5)/5=+-(x+2)/2$

graph{((x+2)^2/4-(y-5)^2/25-1)((y-5)/5-(x+2)/2)((y-5)/5+(x+2)/2)=0 [-12.25, 7.75, 0.2, 10.2]}

How do you find all the critical points to graph (x + 2)^2/4 - (y - 5)^2/25 = 1 (x+2)24−(y−5)225=1(x + 2)^2/4 - (y - 5)^2/25 = 1 including vertices, foci and asymptotes?