Verified answer

Start out by grouping the x and y terms together to get:

(16x^2 + 64x) + (-9y^2 + 72y) - 224 = 0.

Factoring out the x^2 and y^2 coefficient from the first and second set of terms respectively gives:

16(x^2 + 4x) - 9(y^2 - 8y) - 224 = 0.

Completing the square:

16(x^2 + 4x + 4) - 9(y^2 - 8y + 16) - 224 = 16(4) - 9(16)

==> 16(x + 2)^2 - 9(y - 4)^2 - 224 = -80

==> 16(x + 2)^2 - 9(y - 4)^2 = 144

==> (x + 2)^2/9 - (y - 4)^2/16 = 1, by dividing both sides by 144.

By comparing this equation to standard form:

(x - h)^2/a^2 - (y - k)^2/b^2 = 1,

we see that the center of the hyperbola is located at (h, k) = (-2, 4) and:

a^2 = 9 and b^2 = 16.

With c^2 = a^2 + b^2:

c^2 = 9 + 16 = 25 ==> c = 5.

The foci are located c units to the left and to the right of the center. Since c = 5 and the center is located at (-2, 4), the foci are located at (-2 - 5, 4) = (-7, 4) and (-2 + 5, 4) = (3, 4).

I hope this helps!

16x^2–9y^2+64x+72y–224=0

16(x² + 4x + 4) - 64 - 9(y² - 8y + 16) + 144 - 224 = 0

16(x+2)² - 9(y - 4)² = 144

(x+2)²/3² - (y-4)²/4² = 1

Now you know the center, conjugate and transverse axes, use the characteristics of the hyperbola to determine the foci: a² + b² = c² = 25 ==> c = ±5

Since the hyperbola is horizontal (x² term carries the positive sign), the foci are at (-2±5, 4)

9y^2 - 72y - 16x^2 - 64x = sixty 4 9(y^2 - 8y + sixteen) - sixteen(x^2 + 4x + 4) = sixty 4 + a hundred and forty four - sixty 4 9(y - 4)^2 - sixteen(x + 2)^2 = a hundred and forty four (y - 4)^2/sixteen - (x + 2)^2/9 = a million a^2 = sixteen a = 4 b^2 = 9 b = 3 c^2 = a^2 + b^2 = sixteen + 9 = 25 c = 5 center (-2, 4) (-2, 4 ± 5) (-2, 9) and (-2, -a million) 4

16x^2–9y^2+64x+72y–224=0 first complete the squares for both x and y term

{(4x)^2+2(4x)(8)+8^2}-{(3y)^2-2(3y)(12)+12^2}-224-8^2+12^2=0

(4x+8)^2-(3y-12)^2-144=0

or

(4x+8)^2-(3y-12)^2=144

or

(4x+8)^2- (3y-12)^2 =1

----------- ------------

144 144

or

(x+2)^2 - (y-4)^ 2 =1

---------- -----------

9 16

or

(x+2)^2 - (y-4)^2 = 1

---------- -----------

3^2 4^2

comparing with

(x-h)^2 - (y-k)^2 =1

---------- -----------

a^2 b^2

we have h=-2 and k=4

so center of hyperbola is (h.k) i.e. (-2,4)

then the distance "c" from the center to either focus is found by

c^2 = a^2+ b^2

which gives us c=sqrt(25)=5

now axis of parabola is y in this care coz we have a larger term under (y-4)^2. and focie are (h, c+k) and (h.-c+k) so are (-2, 5+2) and (-2,-5+2).

=(-2,7) and (-2,-3)

hope it will help

It equals 0, which means the equation is useless.