Question

What are the foci of the hyperbola given by the equation 16x2 – 9y2 + 64x + 72y – 224 = 0 ?

(–7, –4) and (3, –4)

(–7, 4) and (3, 4)

(–3, –4) and (7, –4)

(–3, 4) and (7, 4)

Answer 1

$$
16x^2-9y^2+64x+72y-224=0\\
16(x^2-2(2)(x)+ \color{red}{?^2})-9(y^2-2(4)(y)+\color{red}{?^2})-224=0
$$

Answer 2

what numbers there would complete the perfect square trinomial? @pvs285

Answer 3

help plz

Answer 4

i need the answer only plz

Answer 5

if you complete the squares and write the equation of the hyperbola in standard form you get

\[\frac{ (x+2)^{2} }{ 9 } - \frac{ ( y - 4)^{2} }{ 16 } = 1\]

Answer 6

$$
\frac{ (x+2)^{2} }{ 3^2 } - \frac{ ( y - 4)^{2} }{ 4^2 } = 1
$$
distance from the center (h,k) to the foci is
$$
\sqrt{a^2+b^2}
$$

Answer 7

ok...

Answer 8

bear in mind that the fraction with the "x" in it has the POSITIVE sign, not the negative, thus the hyperbola opens horizontally, so you move to the foci from the "h" value of the (h,k) center

Answer 9

$$
\text{so one focus is at}\\
h+\color{blue}{\sqrt{a^2+b^2}}\\
\text{and the other focus at}\\
h-\color{blue}{\sqrt{a^2+b^2}}\\
$$

Answer 10

whats the answer need it quaikly

Answer 11

dunno what's \(\sqrt{a^2+b^2} ?\)

Answer 12

now the center of the hyperbola is at (-2,4) and the hyperbola opens right and left.

Since a = 3 and b= 4 the vertex will be 3 points away from the center. So the vertices are ( -5,4) and ( 1,4)

Now to find the focus we need to do c² = a² + b² = 9 + 16 = 25. C² = 25; C = 5

So the focus is 5 points away from the center.

Foci are ( -7,4) and ( 3,4)