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# The directrix of a parabola is x = 4. Its focus is (2,6)

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The directrix of a parabola is x = 4. Its focus is (2,6)

What is the standard form?

$$x-h=a(y-k)^2$$

Jan 15, 2020

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The directrix of a parabola is x = 4. Its focus is (2,6)
What is the standard form?

$$\text{Let a general point on the parabola (x,y) }$$

1. The distance between $$(x,y)$$ and focus $$(2,6)$$ is $$\sqrt{\left(x-2 \right)^2+ \left(y-6 \right)^2}$$

2. The distance between $$(x,y)$$ and the directrix $$(x=4,y)$$ is $$\sqrt{\left(x-4 \right)^2+ \left(y-y \right)^2} =x-4$$

On the parabola, these distances are equal:

$$\begin{array}{rcll} x-4 &=& \sqrt{\left(x-2 \right)^2+ \left(y-6 \right)^2} \\ \left(x-4 \right)^2 &=& \left(x-2 \right)^2+ \left(y-6 \right)^2 \\ x^2-8x+16 &=& x^2-4x+4 + \left(y-6 \right)^2 \\ -8x+16 &=& -4x+4 + \left(y-6 \right)^2 \\ -4x+12 &=& \left(y-6 \right)^2 \quad | \quad :(-4) \\ x+ \dfrac{12}{-4} &=& \dfrac{ \left(y-6 \right)^2}{-4} \\ \mathbf{ x -3 } &=& \mathbf{ \left(-\dfrac{1}{4}\right) \left(y-6 \right)^2 } \\ \end{array}$$

Jan 15, 2020
edited by heureka  Jan 15, 2020