Year 11 Maths - Specialist Graphing techniques

Parabolas

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Questions

Question 1

58636

Find the equation of the locus of points $P(x,y)$ whose distance to the point $F(0,2)$ is equal to the perpendicular distance to the line with equation $y=-2$

Answer

$x^2=8y$

Worked Solution

$$\begin{aligned}
F P&=\sqrt{(x-0)^{2}+(y-2)^{2}}\\
F D&=y+2\\
\therefore \quad(y+2)^{2}&=x^{2}+(y-2)^{2}\\
y^{2}+4 y+4&=x^{2}+y^{2}-4 y+4\\
\therefore\ x^{2}&=8 y
\end{aligned}$$

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