+0  
 
-1
67
1
avatar+9 

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

What is the standard form?

 

Answer in the form

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

 Jan 15, 2020
 #1
avatar+24093 
+3

The directrix of a parabola is x = 4. Its focus is (2,6)
What is the standard form?
Answer in the form x-h=a(y-k)^2

 

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

 

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:

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

 

laugh

 Jan 15, 2020
edited by heureka  Jan 15, 2020

17 Online Users

avatar