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Using a directrix of y = -3 and a focus of (2, 1), what quadratic function is created?
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Find what relation you have between x and y so that \[ (x-2)^2+(y-1)^2=(y+3)^2 \]
Can you finish it now?
Square both sides and simplify to get \[ -4 - 4 x + x^2 - 8 y =0 \] Solve for y in terms of x
You see the \( y^2\) cancels
Solving gives \[ y=\frac{1}{8} \left(x^2-4 x-4\right)=\frac{x^2}{8}-\frac{x}{2}-\frac{1}{2} \] What is the quadratic?
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f(x)=1/8(x-2)^2-1?
@eliassaab
Yep, thanks
YW
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