Find the angle of intersection of the following curves-i y2-x and x2-y-NCERT EXEMPLAR-ii y-x2 and x2-y2-20iii 2 y2-x3 and y2-32 xiv x2-y2-4 x-1-0 and x2-y2-2 y-9-0v x 2 a 2-y 2 b 2-1 and x2-y2-a bvi x2-4 y2-8 and x2-2 y2-2vii x2-27 y and y2-8 xviii x2-y2-2 x and y2-xix y-4-x2 and y-x2-NCERT EXEMPLAR-

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Byju's Answer

Standard XII

Mathematics

Equation of Tangent at a Point (x,y) in Terms of f'(x)

Find the angl...

Question

Find the angle of intersection of the following curves:

(i) y² = x and x² = y [NCERT EXEMPLAR]
(ii) y = x² and x² + y² = 20
(iii) 2y²= x³ and y² = 32x
(iv) x²+ y² − 4x − 1 = 0 and x² + y² − 2y − 9 = 0
(v) $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$ and x² + y² = ab
(vi) x² + 4y² = 8 and x² − 2y² = 2
(vii) x² = 27y and y² = 8x
(viii) x² + y² = 2x and y² = x
(ix) y = 4 − x² and y = x² [NCERT EXEMPLAR]

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Solution

$(i) Given curves are, y^{2} = x . . . (1) x^{2} = y . . . (2) From these two equations, we get {(x^{2})}^{2} = x \Rightarrow x^{4} - x = 0 \Rightarrow x (x^{3} - 1) = 0 \Rightarrow x = 0 or x = 1 Substituting the values of x in (2) we get, y = 0 or y = 1 ∴ (x, y) = (0, 0) or (1, 1) Differenntiating (1) w.r.t. x, 2 y \frac{d y}{d x} = 1 \Rightarrow \frac{d y}{d x} = \frac{1}{2 y} . . . (3) Differenntiating (2) w.r.t . x, 2 x = \frac{d y}{d x} . . . (4) Case -1: (x, y) = (0, 0) The tangent to curve is parallel to x - axis . Hence, the angle between the tangents to two curve at (0, 0) is a right angle . ∴ θ = \frac{π}{2} Case -2 : (x, y) = (1, 1) From (3) we have, m_{1} = \frac{1}{2} From (4) we have, m_{2} = 2 (1) = 2 Now, \tan θ = |\frac{m_{1} - m_{2}}{1 + m_{1} m_{2}}| = |\frac{\frac{1}{2} - 2}{1 + \frac{1}{2} \times 2}| = \frac{3}{4} \Rightarrow θ = \tan^{- 1} (\frac{3}{4})$

$(i i) Given curves are, y = x^{2} . . . (1) x^{2} + y^{2} = 20 . . . (2) From these two equations we get y + y^{2} = 20 \Rightarrow y^{2} + y - 20 = 0 \Rightarrow (y + 5) (y - 4) = 0 \Rightarrow y = - 5 or y = 4 Substituting the values of y in (1) we get, x^{2} = - 5 or x^{2} = 4 \Rightarrow x = \pm 2 and x^{2} = -5 has no real solution So, (x, y) = (2, 4) or (- 2, 4) Differenntiating (1) w.r.t. x, \frac{d y}{d x} = 2 x . . . (3) Differenntiating (2) w.r.t . x, 2 x + 2 y \frac{d y}{d x} = 0 \Rightarrow \frac{d y}{d x} = \frac{- x}{y} . . . (4) Case -1: (x, y) = (2, 4) From (3) we have, m_{1} = 2 (2) = 4 From (4) we have, m_{2} = \frac{- 2}{4} = \frac{- 1}{2} Now, \tan θ = |\frac{m_{1} - m_{2}}{1 + m_{1} m_{2}}| = |\frac{4 + \frac{1}{2}}{1 + 4 (\frac{- 1}{2})}| = \frac{9}{2} \Rightarrow θ = \tan^{- 1} (\frac{9}{2}) Case -2: (x, y) = (- 2, 4) From (3) we have, m_{1} = 2 (- 2) = - 4 From (4) we have, m_{2} = \frac{2}{4} = \frac{1}{2} Now, \tan θ = |\frac{m_{1} - m_{2}}{1 + m_{1} m_{2}}| = |\frac{- 4 - \frac{1}{2}}{1 - 4 (\frac{1}{2})}| = \frac{9}{2} \Rightarrow θ = \tan^{- 1} (\frac{9}{2})$

$(iii) Given curves are, 2 y^{2} = x^{3} . . . (1) y^{2} = 32 x . . . (2) From these two equations we get 2 (32 x) = x^{3} \Rightarrow 64 x = x^{3} \Rightarrow x (x^{2} - 64) = 0 \Rightarrow x = 0, 8, - 8 Substituting the value of x in (2) we get, y_{1} = 0, 16, - 16 ∴ (x_{1}, y_{1}) = (0, 0), (8, 16) or (8, - 16) Differentiating (1) w.r.t. x, 4 y \frac{d y}{d x} = 3 x^{2} \Rightarrow \frac{d y}{d x} = \frac{3 x^{2}}{4 y} . . . (3) Differenntiating (2) w.r.t. x, 2 y \frac{d y}{d x} = 32 \Rightarrow \frac{d y}{d x} = \frac{16}{y} . . . (4) Case - 1: (x, y) = (0, 0) From (3) we have, m_{1} = \frac{0}{0} ∴ We cannot determine θ in this case. Case - 2 : (x, y) = (8, 16) From (3) we have, m_{1} = \frac{192}{64} = 3 From (4) we have, m_{2} = \frac{16}{16} = 1 Now, \tan θ = |\frac{m_{1} - m_{2}}{1 + m_{1} m_{2}}| = |\frac{3 - 1}{1 + 3}| = \frac{2}{4} = \frac{1}{2} \Rightarrow θ = \tan^{- 1} (\frac{1}{2}) Case- 3 : (x_{1}, y_{1}) = (8, - 16) From (3) we have, m_{1} = \frac{192}{- 64} = - 3 From (4) we have, m_{2} = \frac{16}{- 16} = - 1 Now, \tan θ = |\frac{m_{1} - m_{2}}{1 + m_{1} m_{2}}| = |\frac{- 3 + 1}{1 + 3}| = \frac{2}{4} = \frac{1}{2} \Rightarrow θ = \tan^{- 1} (\frac{1}{2})$

$(i v) Given curves are, x^{2} + y^{2} - 4 x - 1 = 0 . . . (1) x^{2} + y^{2} - 2 y - 9 = 0 . . . (2) From (3) we get x^{2} + y^{2} = 4 x + 1 Substituting this in (2), 4 x + 1 - 2 y - 9 = 0 \Rightarrow 4 x - 2 y = 8 \Rightarrow 2 x - y = 4 \Rightarrow y = 2 x - 4 . . . (3) Substituting this in (1), x^{2} + {(2 x - 4)}^{2} - 4 x - 1 = 0 \Rightarrow x^{2} + 4 x^{2} + 16 - 16 x - 4 x - 1 = 0 \Rightarrow 5 x^{2} - 20 x + 15 = 0 \Rightarrow x^{2} - 4 x + 3 = 0 \Rightarrow (x - 3) (x - 1) = 0 \Rightarrow x = 3 or x = 1 Substituting the values of x in (3), we get, y = 2 or y = - 2 ∴ (x, y) = (3, 2), (1, - 2) Differentiating (1) w.r.t. x, 2 x + 2 y \frac{d y}{d x} - 4 = 0 \Rightarrow \frac{d y}{d x} = \frac{4 - 2 x}{2 y} = \frac{2 - x}{y} . . . (4) Differenntiating (2) w.r.t . x, 2 x + 2 y \frac{d y}{d x} - 2 \frac{d y}{d x} = 0 \Rightarrow \frac{d y}{d x} (2 y - 2) = - 2 x \Rightarrow \frac{d y}{d x} = \frac{2 x}{2 - 2 y} = \frac{x}{1 - y} . . . (5) Case - 1: (x, y) = (3, 2) From (4), we get, m_{1} = \frac{2 - 3}{2} = \frac{- 1}{2} From (5), we get, m_{2} = \frac{3}{1 - 2} = - 3 Now, \tan θ = |\frac{m_{1} - m_{2}}{1 + m_{1} m_{2}}| = |\frac{\frac{- 1}{2} + 3}{1 + \frac{3}{2}}| = 1 \Rightarrow θ = \tan^{- 1} (1) = \frac{π}{4} Case - 2: (x, y) = (1, - 2) From (4), we get, m_{1} = \frac{2 - 1}{- 2} = \frac{- 1}{2} From (5), we get, m_{2} = \frac{1}{1 + 2} = \frac{1}{3} Now, \tan θ = |\frac{m_{1} - m_{2}}{1 + m_{1} m_{2}}| = |\frac{\frac{- 1}{2} - \frac{1}{3}}{1 - \frac{1}{6}}| = 1 \Rightarrow θ = \tan^{- 1} (1) = \frac{π}{4}$

$(v) Given curves are, \frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1 . . . (1) x^{2} + y^{2} = a b . . . (2) Multiplying (2) by \frac{1}{a^{2}}, \frac{x^{2}}{a^{2}} + \frac{y^{2}}{a^{2}} = \frac{b}{a} . . . (3) Subtracting (1) from (3), we get \frac{y^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = \frac{b}{a} - 1 \Rightarrow y^{2} (\frac{b^{2} - a^{2}}{a^{2} b^{2}}) = \frac{b - a}{a} \Rightarrow y^{2} = \frac{b - a}{a} \times \frac{a^{2} b^{2}}{(b + a) (b - a)} = \frac{a b^{2}}{b + a} \Rightarrow y = \pm b \sqrt{\frac{a}{b + a}} Substituting this in (3), \frac{x^{2}}{a^{2}} + \frac{a b^{2}}{(b + a) (a^{2})} = \frac{b}{a} \Rightarrow (a + b) x^{2} + a b^{2} = a b^{2} + a^{2} b \Rightarrow x^{2} = \frac{a^{2} b}{a + b} \Rightarrow x = \pm a \sqrt{\frac{b}{a + b}} ∴ (x, y) = (\pm a \sqrt{\frac{b}{a + b}}, \pm b \sqrt{\frac{a}{b + a}}) Now, (x, y) = (a \sqrt{\frac{b}{a + b}}, b \sqrt{\frac{a}{b + a}}) Differentiating (1) w.r.t. x, we get, \frac{2 x}{a^{2}} + \frac{2 y}{b^{2}} \frac{d y}{d x} = 0 \Rightarrow \frac{d y}{d x} = \frac{- x b^{2}}{a^{2} y} \Rightarrow m_{1} = \frac{- a b^{2} \sqrt{\frac{b}{a + b}}}{a^{2} b \sqrt{\frac{a}{b + a}}} = \frac{- b \sqrt{b}}{a \sqrt{a}} Differenntiating (2) w.r.t . x, we get, 2 x + 2 y \frac{d y}{d x} = 0 \Rightarrow \frac{d y}{d x} = \frac{- x}{y} \Rightarrow m_{2} = \frac{- a \sqrt{\frac{b}{a + b}}}{b \sqrt{\frac{a}{b + a}}} = \frac{- a \sqrt{b}}{b \sqrt{a}} We have, \tan θ = |\frac{m_{1} - m_{2}}{1 + m_{1} m_{2}}| = |\frac{\frac{- b \sqrt{b}}{a \sqrt{a}} + \frac{a \sqrt{b}}{b \sqrt{a}}}{1 + (\frac{b \sqrt{b}}{a \sqrt{a}}) (\frac{a \sqrt{b}}{b \sqrt{a}})}| = \frac{\frac{- b^{2} \sqrt{a b} + a^{2} \sqrt{a b}}{a^{2} b}}{\frac{a^{2} b + a b^{2}}{a^{2} b}} = \frac{\sqrt{a b} (a + b) (a - b)}{a^{2} b} \times \frac{a^{2} b}{a b (a + b)} = \frac{a - b}{\sqrt{a b}} \Rightarrow θ = \tan^{- 1} (\frac{a - b}{\sqrt{a b}}) {Similarly, we can prove that θ=tan}^{- 1} (\frac{a - b}{\sqrt{a b}}) for all possibilities of (x, y)$

$(v i) Given curves are, x^{2} + 4 y^{2} = 8 . . . (1) x^{2} - 2 y^{2} = 2 . . . (2) From (1) and (2) we get 6 y^{2} = 6 \Rightarrow y = 1 or y_{1} = - 1 Substituting the values of y in (1) x = 2, - 2 or x = 2, - 2 So, (x, y) = (2, 1), (2, - 1), (- 2, 1), (- 2, - 1) Differenntiating (1) w.r.t. x, 2 x + 8 y \frac{d y}{d x} = 0 \Rightarrow \frac{d y}{d x} = \frac{- x}{4 y} . . . (3) Differenntiating (2) w.r.t. x, 2 x - 4 y \frac{d y}{d x} = 0 \Rightarrow \frac{d y}{d x} = \frac{x}{2 y} . . . (4) Case -1: (x, y) = (2, 1) From (3), we get, m_{1} = \frac{- 1}{2} From (4), we get, m_{2} = 1 We have, \tan θ = |\frac{m_{1} - m_{2}}{1 + m_{1} m_{2}}| = |\frac{\frac{- 1}{2} - 1}{1 - \frac{1}{2}}| = 3 \Rightarrow θ = \tan^{- 1} (3) Case -2: (x, y) = (2, - 1) From (3), we get, m_{1} = \frac{1}{2} From (4), we get, m_{2} = - 1 We have, \tan θ = |\frac{m_{1} - m_{2}}{1 + m_{1} m_{2}}| = |\frac{\frac{1}{2} + 1}{1 - \frac{1}{2}}| = 3 \Rightarrow θ = \tan^{- 1} (3) Case -3 : (x, y) = (- 2, 1) From (3), we get, m_{1} = \frac{1}{2} From (4), we get, m_{2} = - 1 We have, \tan θ = |\frac{m_{1} - m_{2}}{1 + m_{1} m_{2}}| = |\frac{\frac{1}{2} + 1}{1 - \frac{1}{2}}| = 3 \Rightarrow θ = \tan^{- 1} (3) Case -4 : (x, y) = (- 2, - 1) From (3), we get, m_{1} = \frac{- 1}{2} From (4), we get, m_{2} = 1 We have, \tan θ = |\frac{m_{1} - m_{2}}{1 + m_{1} m_{2}}| = |\frac{\frac{- 1}{2} - 1}{1 - \frac{1}{2}}| = 3 \Rightarrow θ = \tan^{- 1} (3)$

$(vii) Given curves are, x^{2} = 27 y . . . (1) y^{2} = 8 x . . . (2) From (2) we get x = \frac{y^{2}}{8} Substituting this in (1), {(\frac{y^{2}}{8})}^{2} = 27 y \Rightarrow y^{4} = 1728 y \Rightarrow y (y^{3} - 12^{3}) = 0 \Rightarrow y = 0 or y = 12 Substituting the v a l u e s o f y in (2), w e g e t, \Rightarrow x = 0 or x = 18 \Rightarrow (x, y) = (0, 0), (18, 12) Differentiating (1) w.r.t. x, 2 x = 27 \frac{d y}{d x} \Rightarrow \frac{d y}{d x} = \frac{2 x}{27} . . . (3) Differenntiating (2) w.r.t . x, 2 y \frac{d y}{d x} = 8 \Rightarrow \frac{d y}{d x} = \frac{4}{y} . . . (4) Case - 1: (x, y) = (0, 0) From (4) we have, m_{2} is undefined ∴ We cannot find θ Case - 2: (x, y) = (18, 12) From (3) we have, m_{1} = \frac{36}{27} = \frac{4}{3} From (4) we have, m_{2} = \frac{4}{12} = \frac{1}{3} Now, \tan θ = |\frac{m_{1} - m_{2}}{1 + m_{1} m_{2}}| = |\frac{\frac{4}{3} - \frac{1}{3}}{1 + \frac{4}{9}}| = \frac{9}{13} \Rightarrow θ = \tan^{- 1} (\frac{9}{13})$

$(v i i i) Given curves are, x^{2} + y^{2} = 2 x . . . (1) y^{2} = x . . . (2) F rom these two equations we get x^{2} + x = 2 x \Rightarrow x^{2} - x = 0 \Rightarrow x (x - 1) = 0 \Rightarrow x = 0 or x = 1 Substituting the values of x in (2) we get, y = 0 or y =±1 ∴ (x, y) = (0, 0), (1, 1), (1, - 1) Differenntiating (1) w.r.t. x, we get, 2 x + 2 y \frac{d y}{d x} = 2 \Rightarrow \frac{d y}{d x} = \frac{1 - x}{y} . . . (3) Differenntiating (2) w.r.t. x,we get, 2 y \frac{d y}{d x} = 1 \Rightarrow \frac{d y}{d x} = \frac{1}{2 y} . . . (4) Case -1: (x, y) = (0, 0) From (3) we get, m_{1} is undefined . ∴ We can not find θ Case -2: Let (x, y) = (1, 1) From (3) we get, m_{1} = 0 From (4) we get, m_{2} = \frac{1}{2} Now, \tan θ = |\frac{m_{1} - m_{2}}{1 + m_{1} m_{2}}| = |\frac{0 - \frac{1}{2}}{1 + 0}| = \frac{1}{2} \Rightarrow θ = \tan^{- 1} (\frac{1}{2}) Case -3: Let (x, y) = (1, - 1) From (3) we get, m_{1} = 0 From (4) we get, m_{2} = \frac{- 1}{2} Now, \tan θ = |\frac{m_{1} - m_{2}}{1 + m_{1} m_{2}}| = |\frac{0 + \frac{1}{2}}{1}| = \frac{1}{2} \Rightarrow θ = \tan^{- 1} (\frac{1}{2})$

$(ix) Given curves are, y = 4 - x^{2} . . . . . (1) y = x^{2} . . . . . (2) From (1) and (2), we get 4 - x^{2} = x^{2} \Rightarrow 2 x^{2} = 4 \Rightarrow x^{2} = 2 \Rightarrow x = \pm \sqrt{2} Substituting the v a l u e s o f x in (2), w e g e t, \Rightarrow y = 2 \Rightarrow (x, y) = (\sqrt{2},2), (- \sqrt{2}, 2) Differentiating (1) w.r.t. x, \frac{d y}{d x} = - 2 x . . . . . (3) Differentiating (2) w.r.t . x, \frac{d y}{d x} = 2 x . . . . . (4) Case 1: (x, y) = (\sqrt{2}, 2) From (3), we have, m_{1} = - 2 \sqrt{2} From (4) we have, m_{2} = 2 \sqrt{2} Now, \tan θ = |\frac{m_{1} - m_{2}}{1 + m_{1} m_{2}}| = |\frac{- 2 \sqrt{2} - 2 \sqrt{2}}{1 - 8}| = \frac{4 \sqrt{2}}{7} \Rightarrow θ = \tan^{- 1} (\frac{4 \sqrt{2}}{7}) Case 1: (x, y) = (- \sqrt{2}, 2) From (3), we have, m_{1} = 2 \sqrt{2} From (4) we have, m_{2} = - 2 \sqrt{2} Now, \tan θ = |\frac{m_{1} - m_{2}}{1 + m_{1} m_{2}}| = |\frac{2 \sqrt{2} + 2 \sqrt{2}}{1 - 8}| = \frac{4 \sqrt{2}}{7} \Rightarrow θ = \tan^{- 1} (\frac{4 \sqrt{2}}{7})$

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