write the standard form equaton of the conic: 4x^2-y^2-16x-4y-4=0

Question

please help!

Answer 1

Given equation is 4x^2-y^2-16x-4y-4 = 0

4x^2-16x-y^2-4y-4 = 0

Add 16 to each side.

4x^2-16x-y^2-4y-4+16 = 0+16

4x^2-16x+16-y^2-4y-4 =16

4(x^2-4x+4)-(y^2+4y+4) = 16

4(x-2)^2-(y+2)^2 = 16

Divide to each side by 16.

4(x-2)^2/16 - (y+2)^2/16 = 16/16

(x-2)^2/4 - (y-(-2))^2/16 = 1

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

Compare i t to standard form of hyperbola is (x-h)^2/a^2 -(y-k)^2/b^2 = 1

Given equation is hyperbola.

Answer 2

The standard form of hyperbola is .

The equation is .

Write the equation in standard form of hyperbola.

To change the expression into a perfect square, first divide the given equation by 4.

.

To change the expression into a perfect square  add (half the x coefficient)² to each side of the expression.

Here x coefficient = - 4, so, (half the x coefficient)² = (- 4/2)²= 4.

Add 4 and 4 to each side.

.

Compare the above equation with standard form of hyperbola.

Therefore, the given conic is a hyperbola.