Question-131584

Question Number 131584 by mr W last updated on 06/Feb/21

Commented by mr W last updated on 06/Feb/21

f i n d R = ?

Answered by ajfour last updated on 06/Feb/21

P o i n t o f c o n t a c t P (t, t^{2})

\frac{d y}{d x} = 2 t = \tan θ

R \sin θ = t

t^{2} - R \cos θ = R

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

\Rightarrow 2 t^{2} = \sqrt{4 t^{2} + 1} + 1

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

4 t^{4} - 8 t^{2} = 0

\Rightarrow t^{2} = 2

t = \sqrt{2}

R = \frac{t}{2 t} \sqrt{1 + 4 t^{2}} = \sqrt{\frac{1}{4} + 2}

R = \frac{3}{2}

t^{2} + R \cos θ = 2 + \frac{1}{2}

t + R \sin θ = \sqrt{2} + \sqrt{2}

E q . o f c i r c l e s :

x^{2} + {(y - \frac{5}{2})}^{2} = \frac{9}{4}

{(x - 2 \sqrt{2})}^{2} + {(y - \frac{3}{2})}^{2} = \frac{9}{4}

Commented by ajfour last updated on 06/Feb/21

Commented by mr W last updated on 06/Feb/21

t h a n k s s i r!

Facebook Tweet Pin