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Question Number 208755 by efronzo1 last updated on 22/Jun/24

Answered by mr W last updated on 22/Jun/24

Commented by mr W last updated on 22/Jun/24

$$y=-\frac{x^2}{2}+4x=-\frac{{(x-4)}^2}{2}+8$$$$y=-\frac{x^2}{2}+8innewsystem\left(red\right)$$$$touchingpointat(p,q)$$$$q=-\frac{p^2}{2}+8$$$$-\frac{1}{\tan \theta }=\frac{dy}{dx}=-p$$$$\Rightarrow \tan \theta =\frac{1}{p}$$$$p=r\cos \theta =\frac{rp}{\sqrt{1+p^2}}\Rightarrow r=\sqrt{1+p^2}$$$$q=r+r\sin \theta =r(1+\frac{1}{\sqrt{1+p^2}})=-\frac{p^2}{2}+8$$$$\sqrt{1+p^2}(1+\frac{1}{\sqrt{1+p^2}})=-\frac{p^2}{2}+8$$$$2\sqrt{1+p^2}=-p^2+14$$$$p^4-32p^2+192=0$$$$\Rightarrow p^2=16-8=8$$$$r=\sqrt{1+8}=3✓$$

Commented by mr W last updated on 22/Jun/24

Commented by efronzo1 last updated on 23/Jun/24

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