What are the vertex and the focus of (x+y)^2 = 2x - (1/2)?

Use change of variable

u = x + y

v = x - y

Solving for x and y we get:

x = 1/2 (u + v)

y = 1/2 (u - v)

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

u^2 = u + v - 1/2

u^2 - u + 1/4 = v - 1/4

(u - 1/2)^2 = v - 1/4

Vertex = (1/2, 1/4)

4p = 1

p = 1/4

Focus = (1/2, 1/2)

Vertex: u = 1/2, v = 1/4

x = 1/2 (u + v) = 3/8

y = 1/2 (u - v) = 1/8

Focus: u = 1/2, v = 1/2

x = 1/2 (u + v) = 1/2

y = 1/2 (u - v)= 0

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

Focus: (1/2, 0)

Vertex (3/8, 1/8)

Check:

http://www.wolframalpha.com/input/?i=parabola+(x%2...

You're correct, this is a parabola that is neither horizontal nor vertical. It has been rotated by some angle θ.

Start by distributing and moving everything over to one side:

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

x^2 + 2xy + y^2 = 2x - 1/2

x^2 + 2xy + y^2 - 2x + 1/2 = 0

The problem is that xy term. That's how you know it's been rotated. To eliminate it, first find the angle of rotation. I'll spare you the proof and derivation (though you can read it yourself in the source in the link) and just give you the equation:

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

For an equation Ax^2 + Bxy + Cy^2 + Dx + Ey + F = 0,

cot (2θ) = (A - C) / B

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In this case, A = 1, C = 1, and B = 2.

cot (2θ) = (1 - 1) / 2

cot (2θ) = 0

Cotangent is 0 when cosine is 0, so:

2θ = π/2 + kπ

θ = π/4 + kπ/2

θ = π/4, 3π/4, 5π/4, 7π/4

Which θ is the right one? Well, all of them. For example, a vertical positive parabola that's rotated 45 degrees counterclockwise is the same as a horizontal positive parabola that's been rotated 135 degrees.

Simply pick any θ. I'll say θ = π/4. That means the new x+ axis is at 45 degrees, and the new y+ axis is at 135 degrees.

Now, to convert the conic section to our new coordinate system, use the equations below:

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

To convert a conic section P(x,y) to a new coordinate system P(x̂,ŷ):

x = x̂ cos θ - ŷ sin θ

y = x̂ sin θ + ŷ cos θ

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

Since θ = π/4,

cos θ = sin θ = 1/√2

Therefore:

x = x̂ / √2 - ŷ / √2

y = x̂ / √2 + ŷ / √2

Substitute:

x^2 + 2xy + y^2 - 2x + 1/2 = 0

(x̂/√2 - ŷ/√2)^2 + 2(x̂/√2 - ŷ/√2)(x̂/√2 + ŷ/√2) + (x̂/√2 + ŷ/√2)^2 - 2(x̂/√2 - ŷ/√2) + 1/2 = 0

Simplify:

x̂^2 / 2 - x̂ŷ + ŷ^2 / 2 + 2(x̂^2 / 2 - ŷ^2 / 2) + x̂^2 / 2 + x̂ŷ + ŷ^2 / 2 - 2x̂/√2 + 2ŷ/√2 + 1/2 = 0

Cancel out the x̂ŷ terms:

x̂^2 / 2 + ŷ^2 / 2 + 2(x̂^2 / 2 - ŷ^2 / 2) + x̂^2 / 2 + ŷ^2 / 2 - 2x̂/√2 + 2ŷ/√2 + 1/2 = 0

Simplify further:

x̂^2 + ŷ^2 + x̂^2 - ŷ^2 - 2x̂/√2 + 2ŷ/√2 + 1/2 = 0

2 x̂^2 - 2x̂/√2 + 2ŷ/√2 + 1/2 = 0

Now complete the square. Start by dividing both sides by 2 to make the leading coefficient 1:

x̂^2 - x̂/√2 + ŷ/√2 + 1/4 = 0

Take half of the x̂ coefficient, square it, and add it to both sides. (1 / (2√2))^2 = (1 / √8)^2 = 1/8.

x̂^2 - x̂/√2 + 1/8 + ŷ/√2 + 1/4 = 1/8

Factor the perfect square:

(x̂ - 1/√8)^2 + ŷ/√2 + 1/4 = 1/8

Write in standard form:

(x̂ - 1/√8)^2 + ŷ/√2 = -1/8

(x̂ - 1/√8)^2 = -ŷ/√2 - 1/8

(x̂ - 1/√8)^2 = -1/√2 (ŷ + √2/8)

Here we have a "vertical down" parabola where:

vertex is (1/√8, -√2/8)

focus is (1/√8, -√2/8 - 1/(4√2)) = (1/√8, -√2/4)

But wait, these are still in the x̂ŷ coordinate. How do we convert back to the original xy coordinate system? By using the same equations we used to do the first conversion.

x = x̂ cos θ - ŷ sin θ

y = x̂ sin θ + ŷ cos θ

x = x̂ / √2 - ŷ / √2

y = x̂ / √2 + ŷ / √2

So the vertex is:

(1/√16 + 1/8, 1/√16 - 1/8) = (3/8, 1/8)

And the focus is:

(1/√16 + 1/4, 1/√16 - 1/4) = (1/2, 0)

Fun tool for rotated parabola

https://www.desmos.com/calculator/hguanwbkbu

https://www.sparknotes.com/math/precalc/conicsecti...

Rotate it.