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lols, im confused as well, i've attempted on drawing it but i think that it's wrong =/

ah Okay I think I got it

maybe the figure is like this|dw:1319019992221:dw|

|dw:1319020480071:dw|

|dw:1319020633975:dw|

|dw:1319020681433:dw|

sorry, typo ignore the last expression

\[\frac{x^2 + 2a*4a}{4a}=c\]
\[\frac{x^2 + 8a^2}{4a}=c\]

OM is from origin perpendicular to PK
equation of OM \[y' = \frac{x}{2a}x'\]

equation of PK \[y' = x' \frac{-2a}{x} + \frac{x^2 + 8a^2}{4a} \]

The point M satisfies both the equation i.e equation of PK and equation of OM

\[x_1 \left( \frac{4a^2 + x^2 }{2ax}\right) = \frac{x^2 + 8a^2}{4a}\]

\[x_1 = \frac{x^3 + 8xa^2}{8a^2 + 2x^2}\]

\[y_1 = \frac{x}{2a}\times \frac{x^3 + 8a^2x}{8a^2 +2x^2} \]

I'm lost =/
Isn't the equation of the normal y = -x + 2ap^3/p

It's okay never mind then, I will ask my teacher tomorrow.