Sketch the following conic sections:a) y^2-4y+8-4x^2=8 b)y^2-4y-4x^2=8 c)3(y-5)^2-7(x+1)^2=1

The answer is given below in the photo

a) 

Solve for y:
${\displaystyle y^2-4y-4x^2+8=0}$

Hint: Solve the quadratic equation by completing the square.
Subtract ${\displaystyle -4x^2+8}$ from both sides:
${\displaystyle y^2-4y=4x^2-8}$

Hint: Take one half of the coefficient of y and square it, then add it to both sides.
Add 4 to both sides:
${\displaystyle y^2-4y+4=4x^2-4}$

Hint: Factor the left hand side.
Write the left hand side as a square:
$(y-2{)}^2=4x^2-4$

Hint: Eliminate the exponent on the left hand side.
Take the square root of both sides:
${\displaystyle y-2=\sqrt{4x^2-4}\text{or}y-2=-\sqrt{4x^2-4}}$

Hint: Look at the first equation: Solve for y.
Add 2 to both sides:
${\displaystyle y=\sqrt{4x^2-4}+2\text{or}y-2=-\sqrt{4x^2-4}}$

Hint: Look at the second equation: Solve for y.
Add 2 to both sides:
Answer: ${\displaystyle y=\sqrt{4x^2-4}+2\text{or}y=2-\sqrt{4x^2-4}}$

b) 

Solve for y:
${\displaystyle y^2-4y-4x^2=8}$

Hint: Solve the quadratic equation by completing the square.
Add $4x^2$ to both sides:
$y^2-4y=4x^2+8$

Hint: Take one half of the coefficient of y and square it, then add it to both sides.
Add 4 to both sides:
$y^2-4y+4=4x^2+12$

Hint: Factor the left hand side.
Write the left hand side as a square:
$(y-2{)}^2=4x^2+12$

Hint: Eliminate the exponent on the left hand side.
Take the square root of both sides:
${\displaystyle y-2=\sqrt{4x^2+12}\text{or}y-2=-\sqrt{4x^2+12}}$

Hint: Look at the first equation: Solve for y.
Add 2 to both sides:
${\displaystyle y=\sqrt{4x^2+12}+2\text{or}y-2=-\sqrt{4x^2+12}}$

Hint: Look at the second equation: Solve for y.
Add 2 to both sides:
Answer: ${\displaystyle y=\sqrt{4x^2+12}+2\text{or}y=2-\sqrt{4x^2+12}}$

c) 

Solve for y:
${\displaystyle 3{(y-5)}^2-7{(x+1)}^2=1}$

Hint: Isolate terms with y to the left hand side.
Add $7(x+1{)}^2$ to both sides:
${\displaystyle 3{(y-5)}^2=7{(x+1)}^2+1}$

Hint: Divide both sides by a constant to simplify the equation.
Divide both sides by 3:
${\displaystyle {(y-5)}^2=\frac{7}{3}{(x+1)}^2+\frac{1}{3}}$

Hint: Eliminate the exponent on the left hand side.
Take the square root of both sides:
${\displaystyle y-5=\sqrt{\frac{7}{3}{(x+1)}^2+\frac{1}{3}}\text{or}y-5=-\sqrt{\frac{7}{3}{(x+1)}^2+\frac{1}{3}}}$

Hint: | Look at the first equation: Solve for y.
Add 5 to both sides: