Question-208755

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} + 4 x = - \frac{{(x - 4)}^{2}}{2} + 8

y = - \frac{x^{2}}{2} + 8 i n n e w s y s t e m (r e d)

t o u c h i n g p o i n t a t (p, q)

q = - \frac{p^{2}}{2} + 8

- \frac{1}{\tan θ} = \frac{d y}{d x} = - p

\Rightarrow \tan θ = \frac{1}{p}

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

q = r + r \sin θ = 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} - 32 p^{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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