Oxford University MAT 2018 Solved Papers

Question : 7 of 10

Marks: +1, -0

The parabolas with equations

y=x^2+c

and

y^2=x

touch (that is, meet tangentially) at a single point. It follows that

c

equals

Solution:

The curves meet and have the same gradient, so

x^2+c \;\; =x^{1 \/ 2}

2 x \;\; =\;{1}/{2 x^{1 \/ 2}}

The second equation gives

x=4^{-2 \/ 3}

, and then substituting this into the first equation gives

c=4^{-1 \/ 3}-4^{-4 \/ 3}=4^{-1 \/ 3}(1-\;{1}/{4})=\;{3}/{4 \;^{3}√{4}}

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