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Question Number 161296 by amin96 last updated on 15/Dec/21

Answered by mr W last updated on 16/Dec/21

$$A(p^2,p)$$$$C(q^2,q)$$$$\tan \theta =\frac{dy}{dx}=\frac{1}{2\sqrt{x}}=\frac{1}{2p}=1$$$$\Rightarrow p=\frac{1}{2}$$$$BC=q-p=\frac{AB}{2}=\frac{q^2-p^2}{2}$$$$\frac{q+p}{2}=1$$$$\Rightarrow q=2-p=\frac{3}{2}$$$$BC=q-p=\frac{3}{2}-\frac{1}{2}=1$$

Commented by mr W last updated on 15/Dec/21

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