FP1 2016 9 Edexcel

9

a) xy=25 \Rightarrow y=\frac{25}{x} \Rightarrow \frac{\mathrm{d}y }{\mathrm{d} x}=\frac{-25}{x^2}
Gradient at P = \frac{-25}{(5p)^2}=\frac{-1}{p^2}
\therefore gradient of normal at P = p^2
Y=mX+c \Rightarrow \frac{5}{p}=p^2(5p)+c
y=p^2x+\frac{5}{p}-5p^3
y-p^2x=\frac{5}{p}-5p^3

b) let y=-x in previous equation
-x-p^2x=\frac{5}{p}-5p^3
x(1+p^2)= -\frac{5}{p}+5p^3
x=\frac{5p^3-\frac{5}{p}}{1+p^2}
=\frac{5p^4-5}{p(1+p^2)}
=\frac{5p^2-5}{p}
=5p-\frac{5}{p}
And y=-x = \frac{5}{p}-5p

c) M=(\frac{5p+(5p-5/p)}{2},\frac{5/p+(5/p-5p)}{2})
=(5p-\frac{5}{2p},\frac{5}{p}-\frac{5p}{2})

We know M is on the x-axis \therefore the y co-ordinate is 0
\frac{5}{p}-\frac{5p}{2}=0
p^2=2
p=\pm\sqrt{2}
Which gives x=\pm\frac{15\sqrt{2}}{4}
But x>0 so x=\frac{15\sqrt{2}}{4}

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:exploding_head: :grin:

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