Question-180104

Question Number 180104 by mr W last updated on 07/Nov/22

Commented by mr W last updated on 07/Nov/22

A l i e s o n x - a x i s a n d B o n t h e c u r v e

y = x^{2} . P i s a t (4, 4) . f i n d t h e s m a l l e s t

p e r i m e t e r o f t r i a n g l e P A B .

Answered by mr W last updated on 07/Nov/22

Commented by mr W last updated on 07/Nov/22

s a y B (t, t^{2})

t a n g e n t a t B :

y - t^{2} = 2 t (x - t)

2 t x - y - t^{2} = 0

P_{1} (4, - 4)

P_{2} (x_{2}, y_{2}) = i m a g e o f P i n t a n g e n t

x_{2} = 4 - \frac{4 t (8 t - 4 - t^{2})}{4 t^{2} + 1}

y_{2} = 4 + \frac{2 (8 t - 4 - t^{2})}{4 t^{2} + 1}

P_{2}, B a n d P_{1} s h o u l d b e c o l l i n e a r,

\frac{x_{2} - t}{t - 4} = \frac{y_{2} - t^{2}}{t^{2} + 4}

\frac{(4 - t) (4 t^{2} + 1) - 4 t (8 t - 4 - t^{2})}{t - 4} = \frac{(4 - t^{2}) (4 t^{2} + 1) + 2 (8 t - 4 - t^{2})}{t^{2} + 4}

2 t^{4} - 16 t^{3} + t^{2} - 12 t + 64 = 0

\Rightarrow t \approx 1.5475

P = \sqrt{{(t - 4)}^{2} + {(t^{2} - 4)}^{2}} + \sqrt{{(t - 4)}^{2} + {(t^{2} + 4)}^{2}}

P_{m i n} \approx 9.7801

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