Find-the-angle-between-the-curves-1-x-2-y-1-y-and-x-3-2-2y-2-x-2-y-2-a-2-2-and-x-2-y-2-a-2-

Question Number 58097 by rahul 19 last updated on 17/Apr/19

F i n d t h e a n g l e b e t w e e n t h e c u r v e s :

1) x^{2} y = 1 - y a n d x^{3} = 2 - 2 y .

2) x^{2} + y^{2} = a^{2} \sqrt{2} a n d x^{2} - y^{2} = a^{2} .

Answered by tanmay last updated on 17/Apr/19

1) x^{2} y = 1 - y

y = \frac{1}{1 + x^{2}}

x^{3} = 2 - 2 y

x^{3} = 2 - 2 (\frac{1}{1 + x^{2}})

x^{3} (1 + x^{2}) = 2 (1 + x^{2}) - 2

x^{3} + x^{5} = 2 x^{2}

x^{2} (x + x^{3} - 2) = 0

x^{2} {(x - 1) + x^{3} - 1} = 0

x^{2} (x - 1) (1 + x^{2} + x + 1) = 0

x^{2} (x - 1) (x^{2} + x + 2) = 0

x = 0

x = 1

x = \frac{- 1 \pm \sqrt{1 - 4 \times 1 \times 2}}{2 \times 1}

x = \frac{- 1 \pm \sqrt{7} i}{2} \leftarrow n o t f e a s i b l e

f i r s t c u r v e y = \frac{1}{1 + x^{2}}

\frac{d y}{d x} = \frac{- 1}{{(1 + x^{2})}^{2}} \times 2 x

m_{1} = \frac{- 2 \times 1}{{(1 + 1)}^{2}} = \frac{- 1}{2}

2 y = 2 - x^{3}

y = 1 - \frac{x^{3}}{2} \to \frac{d y}{d x} = \frac{- 3}{2} \times x^{2} = \frac{- 3 x^{2}}{2}

w h e n p o i n t (x = 1)

m_{2} = \frac{- 3}{2} [p o i n t x = 1]

t a n θ_{2} = \frac{\frac{- 1}{2} + \frac{3}{2}}{1 + (\frac{- 1}{2}) (\frac{- 3}{2})}

= \frac{\frac{2}{2}}{1 + \frac{3}{4}} = \frac{4}{7} \to θ_{2} = {t a n}^{- 1} (\frac{4}{7})

d i s c u s s i o n \dots

l e t t w o c u r v e i n t e s e c t a t p o i n t (x_{1}, y_{1})

t h e n a n g l e b e t w e e n t w o t a n g e n t s a t t h a t

p o i n t (x_{1}, y_{1}) i s t h e a n g l e b e t w e e n t w o c u r v e s

Commented by rahul 19 last updated on 17/Apr/19

T h a n k u s i r!

Commented by tanmay last updated on 18/Apr/19

m o s t w e l c o m e r a h u l \dots

Answered by tanmay last updated on 17/Apr/19

2) x^{2} + y^{2} = a^{2} \sqrt{2}

x^{2} - y^{2} = a^{2}

2 x^{2} = a^{2} (1 + \sqrt{2})

x^{2} = \frac{a^{2} (1 + \sqrt{2})}{2} \to x = a \times \sqrt{\frac{\sqrt{2} + 1}{2}}

y^{2} = a^{2} \sqrt{2} - \frac{a^{2} (1 + \sqrt{2})}{2}

y^{2} = \frac{2 \sqrt{2} a^{2} - a^{2} - \sqrt{2} a^{2}}{2}

y^{2} = \frac{a^{2} (\sqrt{2} - 1)}{2} \to y = a \sqrt{\frac{\sqrt{2} - 1}{2}}

f i r s t c u r v e

x^{2} + y^{2} = a^{2} \sqrt{2}

2 x + 2 y \times \frac{d y}{d x} = 0

\frac{d y}{d x} = \frac{- x}{y} \to \frac{- 1 \times (a \times \sqrt{\frac{\sqrt{2} + 1}{2}})}{a \times \sqrt{\frac{\sqrt{2} - 1}{2}}}

m_{1} = - 1 \times \sqrt{\frac{\sqrt{2} + 1}{\sqrt{2} - 1}} = - 1 \times (\frac{\sqrt{2} + 1}{1})

s e c o n d c u r v e . .

x^{2} - y^{2} = a^{2}

y^{2} = x^{2} - a^{2}

y^{2} = a^{2} (\frac{1 + \sqrt{2}}{2}) - a^{2}

y^{2} = \frac{a^{2} (\sqrt{2} - 1)}{2}

n o w x^{2} - y^{2} = a^{2}

2 x - 2 y \times \frac{d y}{d x} = 0

\frac{d y}{d x} = \frac{x}{y} \to m_{2} = \frac{a \times \sqrt{\frac{\sqrt{2} + 1}{2}}}{a \sqrt{\frac{\sqrt{2} - 1}{2}}} = \sqrt{2} + 1

t a n θ = \frac{m_{2} - m_{1}}{1 + m_{1} m_{2}} = \frac{2 (\sqrt{2} + 1)}{1 + (\sqrt{2} + 1) \times {- 1 (\sqrt{2} + 1)}}

t s n θ = \frac{2 (\sqrt{2} + 1)}{1 - (3 + 2 \sqrt{2})} = \frac{2 (\sqrt{2} + 1)}{1 - 3 - 2 \sqrt{2}} = \frac{2 (\sqrt{2} + 1)}{- 2 (1 + \sqrt{2})} = - 1

t a n θ = - 1 = t a n 135^{o}

θ = 135^{o}

Commented by rahul 19 last updated on 17/Apr/19

Ans given is 45°.

Commented by MJS last updated on 17/Apr/19

without going through above workings :

the angle between 2 curves = the angle

between their tangents . the angle between

2 straight lines is not unique, {it}^{'} s either

α or 180 ° - α \Rightarrow 45 ° is ok and 135 ° is ok too

Commented by tanmay last updated on 18/Apr/19

t h a n k y o u s i r \dots

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