Circle (Ellipse) Graph Question?

Question

Find the coordinates of the points of intersection of y=x/2 with
a) x(square) - y(square) =1
b) x(square)/4 + y(square) =1

Answer 1

a) x² - y² =1
This is the equation of a pair of hyperbolas.

x² - y² = 1
x² - (x/2)² = 1
x² - x²/4 = 1
3x²/4 = 1
x² = 4/3
x = ±2/√3
y = x/2 = ±1/√3

The intersections are (2/√3,1/√3) and (-2/√3,-1/√3).

a) x²/4 + y² = 1
This is the equation of an ellipse.

x²/4 + y² = 1
x²/4 + (x/2)² = 1
x²/4 + x²/4 = 1
x²/2 = 1
x² = 2
x = ±√2
y = x/2 = ±1/√2

The intersections are (√2,1/√2) and (-√2,-1/√2).

Answer 2

y = x/2
a)
for the circle
x^2 + y^2 = 1
x^2 + x^2/4 = 1
5x^2/4 = 1
5x^2 = 4
x^2 = 4/5 = 4*5/25
x = (2/5)√5, -(2/5)√5
y = (1/5)√5, -(1/5)√5

for x^2 - y^2 = 1
x^2 - x^2/4 = 1
3x^2/4 = 1
x^2 = 4*3/9
x = (2/3)√3, -(2/3)√3
y = (1/3)√3, -(1/3)√3

b)
x^2/4 + y^2 = 1
x^2/4 + x^2/4 = 1
x^2/2 = 1
x^2 = 2
x = √2, -√2
y = (1/2)√2, -(1/2)√2

Answer 3

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

x1=-2/sqrt(3)
y1=-1/sqrt(3)


x2=2/sqrt(3)
y2=1/sqrt(3)

Solve b in the same way.