Plot the curve y 2 = x - 1 x + 1 .

y 2 = x - 1 x + 1 i) For Domain y = ± x - 1 x + 1 x - 1 x + 1 ≥ 0 Domain: x ∈ ( - ∞ , - 1 ) ∪ [ 1 , ∞ ) ii) Replace y by - y . Curve remains the same. ⇒ The curve is symmetrical about x -axis. iii) If y = 0 ⇒ x = 1 curve meets x -axis at ( 1 , 0 ) but it does not meet y -axis. iv) It does not pass through the origin. v) Asymptote y 2 ( x + 1 ) = x - 1 Vertical asymptote, coefficient of y 2 = 0 ⇒ x + 1 = 0 x = - 1 y 2 x + y 2 - x + 1 = 0 x ( y 2 - 1 ) + y 2 + 1 = 0 Horizontal asymptote, coefficient of x = 0 ⇒ y 2 - 1 = 0 ⇒ y = ± 1 vi) y = ± x - 1 x + 1 = ± 1 - 1 x 1 + 1 x If x → ∞ , y → ± 1 If x → - ∞ , y → ± 1 For x > 1 , - 1 < y < 1 For x < - 1 , y ∈ ( - ∞ , - 1 ) ∪ ( 1 , ∞ ) So the Curve

Plot the curve: y 2 = x 2 1 - x 1 + x 2

Given: y 2 = x 2 1 - x 1 + x 2 (i) Since, y 2 ≥ 0 ⇒ x 2 1 - x 1 + x 2 ≥ 0 ⇒ x ∈ ( - ∞ , 1 ] ∪ 0 - { - 1 } (ii) At x = 0 , 1 ⇒ y = 0 Also, x = 0 is a repeated root, so graph will touch x -axis at ( 0 , 0 ) As x → ∞ , y → - ∞ As x → - ∞ , y → ∞ (iii) Consider, y = f ( x ) = x 2 ( 1 - x ) ( 1 + x ) 2 Asymptote: (a) As x → - 1 , y → ∞ So, x = - 1 is a vertical asymptote (b) There is no horizontal asymptote (c) For oblique asymptote, Consider x 2 ( 1 - x ) ( 1 + x ) 2 ( x + 1 ) 2 x 2 ( 1 - x ) ⇒ ( x + 1 ) 2 - x + 3 - x 3 + x 2 - x 3 - 2 x 2 - x 3 x 2 + x 3 x 2 + 6 x + 3 - 5 x - 3 ⇒ y = 3 - x + - 5 x - 3 x + 1 2 So, y = - x + 3 is an oblique asymptote (iv) But x ∈ ( - ∞ , 1 ] - { - 1 } So we have to consider the graph of y = f ( x ) = x 2 ( 1 - x ) ( 1 + x ) 2 only for x ≤ 1 (v) For graph of y 2 = x 2 ( 1 - x ) ( 1 + x ) 2 Replace y → - y So, the graph is symmetric about x -axis. We have to take image of graph of y = f ( x ) in x -axis, (along with original graph of y = f ( x ) )

Plot the curve x = 1 4 t + 1 2 , y = 1 4 t - 1 2 .

Given, x = 1 4 ( t + 1 ) 2 , y = 1 4 ( t - 1 ) 2 Here x ≥ 0 , ∀ t ∈ R y ≥ 0 , ∀ t ∈ R Now 4 x = t 2 + 2 t + 1 … i 4 y = t 2 - 2 t + 1 … ii using equation ( i ) and ( ii ) 4 x - 4 y = 4 t ⇒ t = x - y … iii Substituting the value of t 4 x = ( x - y + 1 ) 2 ∵ x = 1 4 ( t + 1 ) 2 x 2 + y 2 + 1 - 2 x y - 2 y + 2 x = 4 x x 2 + y 2 - 2 x y - 2 y - 2 x + 1 = 0 … iv It's a second-degree equation in x and y Comparing above equation with a x 2 + b y 2 + 2 h x y + 2 g x + 2 f y + c = 0 a = 1 , b = 1 , h = - 1 , g = - 1 , f = - 1 , c = 1 Now △ = a h g h b f g f c = 1 - 1 - 1 - 1 1 - 1 - 1 - 1 1 △ = - 4 Here △ ≠ 0 ⇒ It will represent a conic Now h 2 - a b = 1 - 1 = 0 ⇒ It will represent a parabola i) Replace y by x in equation ( iv ) . The curve remains the same. ⇒ The curve is symmetrical about the line y = x . ⇒ The axis of the parabola is y = x . Now, The parabola does not pass through the origin. If x = 0 ⇒ y 2 - 2 y + 1 = 0 ⇒ ( y - 1 ) 2 = 0 ⇒ y = 1 If y = 0 ⇒ x 2 - 2 x + 1 = 0 ⇒ ( x - 1 ) 2 = 0 ⇒ x = 1 It meets(touches) the x -axis at ( 1 , 0 ) and y -axis at ( 0 , 1 ) Let S = x 2 + y 2 - 2 x y - 2 x - 2 y + 1 Let point A ( - 1 , - 1 ) ( 3 r d quadrant) S 1 = 1 + 1 - 2 + 2 + 2 + 1 = 5 S 1 > 0 ⇒ (For any point x 1 , y 1 which lies in 3 rd quadrant S 1 > 0 )The point A lies outside to the parabola. ⇒ Parabola is not there in 3 rd quadrant. It's only in 1 st quadrant ( x ≥ 0 , y ≥ 0 )

Plot the curve: x = t 2 1 - t 2 , y = 1 1 + t 2

Given: x = t 2 1 - t 2 , y = 1 1 + t 2 i) Here x = t 2 1 - t 2 , t ≠ ± 1 ⇒ t 2 = x 1 + x We know t 2 ≥ 0 x 1 + x ≥ 0 x ∈ ( - ∞ , - 1 ) ∪ [ 0 , ∞ ) So the domain of y = f ( x ) ∈ ( - ∞ , - 1 ) ∪ [ 0 , ∞ ) ii) Now as t ≠ ± 1 y = 1 1 + t 2 ⇒ y ≠ 1 2 here y > 0 ∀ t ∈ R and 1 1 + t 2 ≤ 1 ∀ t ∈ R ⇒ 0 < y ≤ 1 but y ≠ 1 2 ⇒ y ∈ 0 , 1 2 ∪ ( 1 2 , 1 ] iii) x = t 2 1 - t 2 , y = 1 1 + t 2 t 2 = x 1 + x ⇒ y = 1 1 + x 1 + x = 1 + x 1 + 2 x So, y = x + 1 2 x + 1 lim x → ∞ x + 1 2 x + 1 = lim x → ∞ x 1 + 1 x x 2 + 1 x = 1 2 ⇒ If x → ∞ , y → 1 2 If x → - ∞ , y → 1 2 iv) y = x + 1 2 x + 1 at x = 0 , y = 1 It meets at ( 0 , 1 ) on y -axis as x → - 1 , y → 0 So the graph of the curve

E M B I B E

Skills in Mathematics Play with Graphs for JEE Main & Advanced>Chapter 3 - Asymptotes, Singular Points and Curve Tracing>EXERCISE

Amit M Agarwal Solutions for Chapter: Asymptotes, Singular Points and Curve Tracing, Exercise 1: EXERCISE

Author:Amit M Agarwal

Amit M Agarwal Mathematics Solutions for Exercise - Amit M Agarwal Solutions for Chapter: Asymptotes, Singular Points and Curve Tracing, Exercise 1: EXERCISE

Attempt the practice questions on Chapter 3: Asymptotes, Singular Points and Curve Tracing, Exercise 1: EXERCISE with hints and solutions to strengthen your understanding. Skills in Mathematics Play with Graphs for JEE Main & Advanced solutions are prepared by Experienced Embibe Experts.