Question-209041

Question Number 209041 by alcohol last updated on 30/Jun/24

Commented by alcohol last updated on 30/Jun/24

f i n d r

Commented by mr W last updated on 01/Jul/24

Answered by mr W last updated on 01/Jul/24

s a y t o u c h i n g p o i n t o n y = x^{2} a t P (p, p^{2})

c e n t e r o f c i r c l e C (2 - r, 2 - r)

\tan θ = 2 p

2 - r = p + r \sin θ \Rightarrow 2 - r = p + \frac{2 p r}{\sqrt{1 + 4 p^{2}}}

\Rightarrow r = \frac{2 - p}{1 + \frac{2 p}{\sqrt{1 + 4 p^{2}}}}

2 - r = p^{2} - r \cos θ \Rightarrow 2 - r = p^{2} - \frac{r}{\sqrt{1 + 4 p^{2}}}

\Rightarrow r = \frac{2 - p^{2}}{1 - \frac{1}{\sqrt{1 + 4 p^{2}}}}

\frac{2 - p^{2}}{1 - \frac{1}{\sqrt{1 + 4 p^{2}}}} = \frac{2 - p}{1 + \frac{2 p}{\sqrt{1 + 4 p^{2}}}}

\Rightarrow p \approx 1.3328, r \approx 0.3446

Commented by mr W last updated on 30/Jun/24

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