Find the equation of the parabola, with axis of symmetry of the y-axis, which passes through the points 𝐴(−2,1) and 𝐵(4, −5).

Answer 1

The general form for this equation is below. Since the axis of symmetry is the y axis, there is no h.

x²=a(y-k)

If you plug in the 2 points given, you have 2 equations, 2 unknowns

(-2)²=a(1-k) and (4)²=a(-5-k)

Simplify

4=a(1-k) and 16=a(-5-k)

You can solve this by substitution. Solve the first equation for a: a=4/(1-k) and plug that into the second equation

16=[4/(1-k)](-5-k) Multiply both sides on the equation by (1-k) to get rid of the fraction.

16(1-k)=4(-5-k) Distribute

16-16k=-20-4k Add 16k to both sides

16=-20+12k Add 20 to both sides

36=12k Divide both sides by 12

k=3

Substitute k into the 1st equation.

4=a(1-3)

4=-2a Divide both sides by -2

a=-2

The final equation would be:

x² = -2(y-3)