1.Tac Locus and Node Locus B.sc. Maths

Tac Locus and Node Locus B.sc. Maths

In this article “Tac Locus and Node Locus B.sc. Maths”, we will read to find out the tac-locus,node-Locus and singular solution of differential equations of first order but not of first degree.
“Before moving to Part 2,read our basic introduction to Singular Solutions and Extraneous Loci.”

3.Tac Locus and Node Locus B.sc. Maths Examples

Investigate for singular solution and extraneous loci of the following differential equations:
Example:8. 9p^2(2-y)^2 = 4(3-y)
Solution: 9p^2(2-y)^2 = 4(3-y) \dots(1) \\ \Rightarrow 3p = \pm \frac{2\sqrt{3-y}}{2-y} \\ \Rightarrow 3\frac{dy}{dx} = \pm \frac{2\sqrt{3-y}}{2-y} \\ \Rightarrow \frac{1}{3} \cdot \frac{dx}{dy} = \pm \frac{2-y}{2\sqrt{3-y}} \\ \Rightarrow = \pm \frac{3-y-1}{2\sqrt{3-y}} \\ = \pm \left( \frac{3-y}{2\sqrt{3-y}} - \frac{1}{2\sqrt{3-y}} \right) \\ \Rightarrow \frac{1}{3} \int dx = \pm \int \left( \frac{\sqrt{3-y}}{2} - \frac{1}{2\sqrt{3-y}} \right) dy \\ \text{Integrating} \\ \Rightarrow \frac{x}{3} = \pm \left( -\frac{1}{3}(3-y)^{3/2} + \sqrt{3-y} \right) + c \\ \Rightarrow \frac{1}{3}x = c \pm \sqrt{3-y} \mp \frac{1}{3}(3-y)^{3/2} \\ \Rightarrow \frac{(x-c)^2}{9} = 3-y + \frac{1}{9}(3-y)^3 - \frac{2}{3}(3-y)^2 \\ = 3-y + \frac{1}{9}(3-y)^2(3-y-6) \\ = 3-y + \frac{1}{9}(3-y)^2(-3-y) \\ = (3-y) \left[ 1 + \frac{1}{9}(3-y)(-3-y) \right] \\ = \frac{(3-y)}{9} [9 -9 - 3y - 3y + y^2] \\ \Rightarrow \frac{1}{9}(x-c)^2 = \frac{1}{9}y^2(3-y) \\ \Rightarrow (x-c)^2 = y^2(3-y) \\ \text{which is complete primitive of the equation } \dots(1) \\ \text{write it as} \\ c^2 - 2cx + x^2 - y^2(3-y) = 0 \dots(2) \\ \text{which is general solution c-discriminant} \\ B^2 - 4AC = 0 \\ (-2x)^2 - 4 \times 1 \times [x^2 - y^2(3-y)] = 0 \\ \Rightarrow 4x^2 - 4x^2 + 4y^2(3-y) = 0 \\ \Rightarrow y^2(3-y) = 0 \dots(3) \\ \text{P-Discriminant from (1)} \\ B^2 - 4AC = 0 \\ 0 - 4 \times 9(2-y)^2 \times [-4(3-y)] = 0 \\ \Rightarrow (2-y)^2(3-y) = 0 \dots(4) \\ \text{common factor occurring once in (3) and (4)} \\ 3-y = 0 \Rightarrow y = 3 \text{ is singular solution} \\ \text{The factor } y = 0 \text{ which occurs twice in c-discriminant and} \\ \text{is not in p-discriminant is node locus.} \\ \text{The factor } 2-y = 0 \text{ which occurs twice in p-discriminant and} \\ \text{is not in c-discriminant is tac-locus.}
Example:9. y^2(1+4p^2)-2pxy-1=0
Solution: y^2(1+4p^2)-2pxy-1=0 \dots(1) \\ \text{This equation can be written as} \\ y^2+4p^2y^2-2pxy-1=0 \\ \Rightarrow 2pxy = y^2+4p^2y^2-1 \\ \Rightarrow x = \frac{y^2+4p^2y^2-1}{2py} = \frac{y}{2p} + 2py - \frac{1}{2py} \\ \text{Differentiating w.r.t. } y, \text{we get} \\ \frac{dx}{dy} = \frac{1}{2p} - \frac{y}{2p^2}\frac{dp}{dy} + 2p + 2y\frac{dp}{dy} + \frac{1}{2py^2} + \frac{1}{2p^2y}\frac{dp}{dy} \\ \frac{1}{p} = \frac{1}{2p} - \frac{y}{2p^2}\frac{dp}{dy} + 2p + 2y\frac{dp}{dy} + \frac{1}{2py^2} + \frac{1}{2p^2y}\frac{dp}{dy} \quad [\because \frac{dx}{dy} = \frac{1}{p}] \\ \Rightarrow \frac{1}{2py^2} + \frac{1}{2p} + 2p - \frac{1}{p} + \frac{1}{2p^2y}\frac{dp}{dy} - \frac{y}{2p^2}\frac{dp}{dy} + 2y\frac{dp}{dy} = 0 \\ \Rightarrow \frac{1}{2py^2} + \frac{1+4p^2-2}{2p} + \frac{1}{2p^2y}\frac{dp}{dy} - \frac{y}{2p^2}\frac{dp}{dy} + 2y\frac{dp}{dy} = 0 \\ \Rightarrow \frac{1}{2py^2} - \frac{1}{2p} + 2p + \frac{1}{2p^2y}\frac{dp}{dy} - \frac{y}{2p^2}\frac{dp}{dy} + 2y\frac{dp}{dy} = 0 \\ \Rightarrow \frac{p}{y} \left(\frac{1}{2p^2y} - \frac{y}{2p^2} + 2y\right) + \frac{dp}{dy}\left(\frac{1}{2p^2y} - \frac{y}{2p^2} + 2y\right) = 0 \\ \Rightarrow \left(\frac{p}{y} + \frac{dp}{dy}\right)\left(\frac{1}{2p^2y} - \frac{y}{2p^2} + 2y\right) = 0 \\ \text{Here the first factor when} \\ \text{equated to zero will give primitive} \\ \text{we have} \\ \frac{p}{y} + \frac{dp}{dy} = 0 \\ \frac{dp}{dy} = -\frac{p}{y} \\ \Rightarrow \frac{dp}{p} = -\frac{dy}{y} \\ \text{Integrate:} \\ \log p = -\log y + \log c \\ \Rightarrow p = \frac{c}{y} \dots(2) \\ \text{Eliminating } p \text{ from (1) and (2), we get} \\ y^2\left(1 + \frac{4c^2}{y^2}\right) - 2\frac{c}{y}xy - 1 = 0 \\ \Rightarrow y^2 + 4c^2 - 2cx - 1 = 0 \\ \text{which is general solution} \\ \text{c-discriminant} \\ B^2 - 4AC = 0 \\ (-2x)^2 - 4 \times 4 \times (y^2 - 1) = 0 \\ \Rightarrow 4x^2 - 16(y^2 - 1) = 0 \\ \Rightarrow x^2 - 4y^2 + 4 = 0 \dots(3) \\ \text{p-discriminant} \\ B^2 - 4AC = 0 \\ (-2xy)^2 - 4 \times 4y^2(y^2 - 1) = 0 \\ \Rightarrow 4x^2y^2 - 16y^2(y^2 - 1) = 0 \\ \Rightarrow 4y^2(x^2 - 4y^2 + 4) = 0 \dots(4) \\ \text{common factor occuring once in (3) and (4)} \\ x^2 - 4y^2 + 4 = 0 \text{ is singular solution} \\ \text{The factor } y = 0 \text{ which occurs twice in p-discriminant} \\ \text{and is not in c-discriminant is Tac-locus.}

Tac Locus and Node Locus B.sc. Maths

Example:10. \sin px \cos y - \cos px \sin y - p = 0
Solution: \sin px \cos y - \cos px \sin y - p = 0 \\ \text{This equation can be written as} \\ \sin(px - y) = p \\ \Rightarrow px - y = \sin^{-1} p \\ \Rightarrow y = px - \sin^{-1} p \dots(1) \\ \text{This is Clairaut's form.} \\ \text{Hence replacing } p \text{ by } c, \text{the} \\ \text{general solution is} \\ y = cx - \sin^{-1} c \dots(2) \\ \text{c-discriminant} \\ \text{Differentiating (2) w.r.t. } c, \text{we get} \\ 0 = x - \frac{1}{\sqrt{1-c^2}} \\ \Rightarrow x\sqrt{1-c^2} = 1 \Rightarrow \sqrt{1-c^2} = \frac{1}{x} \\ \Rightarrow 1 - c^2 = \frac{1}{x^2} \Rightarrow c^2 = 1 - \frac{1}{x^2} \\ \Rightarrow c = \sqrt{1 - \frac{1}{x^2}} \\ \text{Put value of } c \text{ in (2), we get} \\ y = x\sqrt{1 - \frac{1}{x^2}} - \sin^{-1}\left(\sqrt{1 - \frac{1}{x^2}}\right) \\ \Rightarrow y = \sqrt{x^2 - 1} - \sin^{-1}\left(\sqrt{1 - \frac{1}{x^2}}\right) \dots(3) \\ \text{p-discriminant} \\ \text{From (1), differentiating} \\ \text{(1) w.r.t. } p \text{ we get} \\ 0 = x - \frac{1}{\sqrt{1-p^2}} \\ \Rightarrow p = \sqrt{1 - \frac{1}{x^2}} \\ \text{Put value of } p \text{ in (1) we get} \\ y = \sqrt{x^2 - 1} - \sin^{-1}\left(\sqrt{1 - \frac{1}{x^2}}\right) \dots(4) \\ \text{common factor occurring once} \\ \text{in (3) and (4)} \\ y = \sqrt{x^2 - 1} - \sin^{-1}\left(\sqrt{1 - \frac{1}{x^2}}\right) \text{ is singular}
Example:11. 4p^2x(x-a)(x-b) = [3x^2-2x(a+b)+ab]^2
Solution: 4p^2x(x-a)(x-b) = [3x^2-2x(a+b)+ab]^2 \dots(1) \\ \text{This equation can be written as} \\ p = \frac{dy}{dx} = \pm \frac{[3x^2-2x(a+b)+ab]}{2\sqrt{x(x-a)(x-b)}} \\ \Rightarrow \frac{dy}{dx} = \pm \frac{[3x^2-2x(a+b)+ab]}{2\sqrt{x^3-x^2(a+b)+abx}} \\ \Rightarrow \int dy = \pm \int \frac{3x^2-2x(a+b)+ab}{2\sqrt{x^3-x^2(a+b)+abx}} dx \\ \text{put } x^3-x^2(a+b)+abx = t \\ \Rightarrow [3x^2-2x(a+b)+ab] dx = dt \\ \Rightarrow y = \pm \int \frac{1}{2\sqrt{t}} dt \\ \Rightarrow y = \pm \sqrt{t} + c \\ \Rightarrow y = c \pm \sqrt{x^3-x^2(a+b)+abx} \\ \Rightarrow (y-c)^2 = x^3-x^2(a+b)+abx \\ \text{Which is complete primitive of} \\ \text{the equation. This can be written} \\ \text{as} \\ c^2 - 2cy + y^2 - x(x-a)(x-b) = 0 \\ \text{c-discriminant} \\ B^2 - 4AC = 0 \\ \Rightarrow (-2y)^2 - 4 \times 1 \times [y^2 - x(x-a)(x-b)] = 0 \\ \Rightarrow 4y^2 - 4[y^2 - x(x-a)(x-b)] = 0 \\ \Rightarrow 4y^2 - 4y^2 + 4x(x-a)(x-b) = 0 \\ \Rightarrow x(x-a)(x-b) = 0 \dots(2) \text{P-discriminant} \\ \text{From (1)} \quad B^2 - 4AC = 0 \\ 0^2 - 4 \times 4x(x-a)(x-b) [3x^2-2x(a+b)+ab]^2 = 0 \\ \Rightarrow x(x-a)(x-b) [3x^2-2x(a+b)+ab]^2 = 0 \dots(3) \\ \text{common factor occurring once} \\ \text{in (2) and (3),} \\ x(x-a)(x-b) \text{ is singular solution.} \\ \text{The factor } 3x^2-2x(a+b)+ab=0 \\ \text{which occurs twice in p-discri-} \\ \text{minant and is not in c-discrimi-} \\ \text{nant is Tac-locus} \\ 3x^2-2x(a+b)+ab=0 \\ \Rightarrow x = \frac{2(a+b) \pm \sqrt{[-2(a+b)]^2 - 4 \times 3 \times ab}}{2 \times 3} \\ = \frac{2(a+b) \pm \sqrt{4(a+b)^2 - 12ab}}{6} \\ = \frac{1}{3}(a+b) \pm \frac{2 \sqrt{a^2+b^2+2ab-3ab}}{6} \\ \Rightarrow x = \frac{1}{3}(a+b) \pm \frac{1}{3} [(a+b)^2 - 3ab]^{1/2}
Example:12. x^2p^2 - 3pxy + 2y^2 + x^3 = 0
Solution: x^2p^2 - 3pxy + 2y^2 + x^3 = 0 \dots(1) \\ \text{Put } x = u, \ \frac{y}{x} = v \\ dx = du, \left( -\frac{y}{x^2} + \frac{1}{x} \frac{dy}{dx} \right) = \frac{dv}{dx} \\ \frac{dv}{du} = P= \frac{1}{(-\frac{y}{x^2} + \frac{1}{x} p)} \\ \Rightarrow P = \frac{1}{-\frac{v}{x} + \frac{p}{x}} = \frac{u}{(-v+p)} \\ \Rightarrow -v+p = \frac{u}{P} \Rightarrow p = \frac{u}{P} + v \\ \text{Put these value in (1), we get} \\ x^2P^2 - 3Px^2 \cdot \frac{y}{x} + 2x^2 \frac{y^2}{x^2} + x^3 = 0 \\ \Rightarrow u^2 \left(\frac{u}{P} + v\right)^2 - 3 \left(\frac{u}{P} + v\right) \cdot u^2v+ 2u^2 \cdot v^2 + u^3 = 0 \\ \Rightarrow u^2 \left( \frac{u^2}{P^2} + u^2 + \frac{2uv}{P} \right) - 3 \\ \left( \frac{u}{P} + v \right) \cdot u^2 \cdot v + 2u^2 v^2 + u^3 = 0 \\ \Rightarrow \frac{u^2}{P^2} + v^2 + \frac{2uv}{P} - \frac{3uv}{P} - 3v^2 + 2v^2 + u = 0 \\ \Rightarrow \frac{u^2}{P^2} - \frac{uv}{P} + u = 0 \\ \Rightarrow \frac{u}{P^2} - \frac{v}{P} + u = 0 \\ \Rightarrow \frac{v}{P} = u + \frac{u}{P^2} \\ \Rightarrow v = uP + \frac{u}{P} \\ \text{This is Clairaut's form.} \\ \text{Hence replacing } p \text{ by } c \text{, the} \\ \text{general equation is} \\ v = xc + \frac{u}{c} \\ \Rightarrow \frac{y}{x} = xc + \frac{x}{c} \\ \Rightarrow y = c x^2 + \frac{x^2}{c} \quad y = cx + \frac{x^2}{c} \\ \Rightarrow c^2 x^2 - cy + x^2 = 0 \Rightarrow c^2 x - cy + x^2 = 0 \\ \text{c-discriminant} \\ B^2 - 4AC = 0 \\ \Rightarrow (-y)^2 - 4 \times x^2 \times x^2 = 0 \\ \Rightarrow y (y^2 - 4x^3) = 0 \dots(2) \\ \text{p-discriminant} \\ \text{From (1), } B^2 - 4AC = 0 \\ (-3xy)^2 - 4 \times x^2 \times (2y^2 + x^3) = 0 \\ \Rightarrow 9x^2 y^2 - 4x^2 (2y^2 + x^3) = 0 \\ \Rightarrow 9x^2 y^2 - 8x^2 y^2 - 4x^5 = 0 \\ \Rightarrow x^2 y^2 - 4x^5 = 0 \\ \Rightarrow x^2 (y^2 - 4x^3) = 0 \dots(3) \\ \text{Common factor occurring once in 2) and (3) }\\ y^2-4x^3=0 \text{is singular solution}
Example:13.Obtain the singular p^2 y^2 \cos^2\alpha - 2pxy \sin^2\alpha + y^2 - x^2 \sin^2\alpha = 0 solution of the equation directly from the equation and also from its complete primitive,explaining the geometrical significance of the irrelevant factors that present themselves.
Solution:The equation is
p^2 y^2 \cos^2\alpha - 2pxy \sin^2\alpha + y^2 - x^2 \sin^2\alpha = 0 \quad \cdots(1) \\ py = \frac{2x \sin^2\alpha \pm \sqrt{(-2x \sin^2\alpha)^2 - 4\cos^2\alpha (y^2 - x^2 \sin^2\alpha)}}{2\cos^2\alpha} \\ = \frac{2x \sin^2\alpha \pm \sqrt{4x^2 \sin^4\alpha + 4x^2 \cos^2\alpha \sin^2\alpha - 4y^2 \cos^2\alpha}}{2\cos^2\alpha} \\ = \frac{2x \sin^2\alpha \pm \sqrt{4x^2 \sin^2\alpha (\sin^2\alpha + \cos^2\alpha) - 4y^2 \cos^2\alpha}}{2\cos^2\alpha} \\ = \frac{2x \sin^2\alpha \pm \sqrt{4x^2 \sin^2\alpha - 4y^2 \cos^2\alpha}}{2\cos^2\alpha} \\ \Rightarrow py = x \tan^2\alpha \pm \sec\alpha \sqrt{x^2 \tan^2\alpha - y^2} \\ \Rightarrow y \frac{dy}{dx} = x \tan^2\alpha \pm \sec\alpha \sqrt{x^2 \tan^2\alpha - y^2} \\ \Rightarrow \frac{y \, dy - x \tan^2\alpha \, dx}{\pm \sqrt{x^2 \tan^2\alpha - y^2}} = \sec\alpha dx \\ \Rightarrow \pm \int \frac{y dy - x \tan^2\alpha \, dx}{\sqrt{x^2 \tan^2\alpha - y^2}} = \int \sec\alpha dx \\ \text{Put } x^2 \tan^2\alpha - y^2 = t \\ \Rightarrow 2x \tan^2\alpha dx - 2y dy = dt \\ \Rightarrow \pm \frac{1}{2} \int \frac{1}{\sqrt{t}} \, dt = c - x \sec\alpha \\ \Rightarrow \pm \sqrt{x^2 \tan^2\alpha - y^2} = c - x \sec\alpha \\ \Rightarrow x^2 \tan^2\alpha - y^2 = (c - x \sec\alpha)^2 \\ \Rightarrow x^2 \tan^2\alpha - y^2 = c^2 - 2cx \sec\alpha + x^2 \sec^2\alpha \\ \Rightarrow c^2 - 2cx \sec\alpha + x^2 + y^2 = 0 \\ \text{Which clearly represents a family of circles for all values of } c. \\ \text{c-discriminant} \\ B^2 - 4AC = 0 \\ \Rightarrow (-2x \sec\alpha)^2 - 4 \times 1 \times (x^2 + y^2) = 0 \\ \Rightarrow 4x^2 \sec^2\alpha - 4x^2 - 4y^2 = 0 \\ \Rightarrow 4x^2 (\sec^2\alpha - 1) - 4y^2 = 0 \\ \Rightarrow x^2 \tan^2\alpha - y^2 = 0 \cdots(2) \\ \text{p-discriminant} \\ \text{From (1), } B^2 - 4AC = 0 \\ (-2xy \sin^2\alpha)^2 = 4y^2 \cos^2\alpha (y^2 - x^2 \sin^2\alpha) \\ \Rightarrow 4x^2 y^2 \sin^4\alpha - 4y^4 \cos^2\alpha + 4x^2 y^2 \cos^2\alpha \sin^2\alpha = 0 \\ \Rightarrow 4x^2 y^2 \sin^2\alpha (\sin^2\alpha + \cos^2\alpha) - 4y^4 \cos^2\alpha = 0 \\ \Rightarrow 4x^2 y^2 \sin^2\alpha - 4y^4 \cos^2\alpha = 0 \\ \Rightarrow 4y^2 \cos^2\alpha (x^2 \tan^2\alpha - y^2) = 0 \\ \Rightarrow y^2 (x^2 \tan^2\alpha - y^2) = 0 \\ \Rightarrow y^2 = 0, \quad y = \pm x \tan\alpha \cdots(3) \\ \text{common factor occurring once in (2) and (3)} y = \pm x \tan\alpha \text{ is singular solution} \\ \text{The factor } y^2 = 0 \text{ which occurs twice in p-discriminant}\\ \text{and is not in c-discriminant is Tac-locus}
With the above illustrations,one can understand the Tac Locus and Node Locus B.sc. Maths.

Tac Locus and Node Locus B.sc. Maths

Also Read This Article:- Locus (mathematics)

3.Practice Questions of Tac Locus and Node Locus B.sc. Maths for Students

Tac Locus and Node Locus B.sc. Maths

Obtain the complete primitive and singular solutions of the following equations:
(1.) y^2 p^2 + y^2 = r^2
(2.) x^3 p^2 + x^2 yp + a^3 = 0
[Hint: put u = \frac{1}{x}, v= y]
Answers: (1.) C.P.: (x + c)^2 + y^2 = r^2,
S.S.: y = \pm r, Tac-locus y = 0
(2.) C.P.: 1 = cxy - a^2 c^2 x,
S.S.: x(xy^2 - 4a^3) = 0, Tac-locus x = 0
By solving the above questions,you can understand the Tac Locus and Node Locus B.sc. Maths well because the concept is well understood when you solve it practically.

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Also Read This Article:-Simultaneous Differential Equations

4.Frequently Asked Questions Related to Tac Locus and Node Locus B.sc. Maths

**छात्र-छात्राओं से आज का सवाल**

*"निम्नलिखित संख्या श्रेणी में एक संख्या असंगत दी गई है,उस असंगत संख्या को ज्ञात करो:"*
0,5,18,43,82,145,230
*"A number in the following number series is given as a non-conformist,find out the odd number:"*
0,5,18,43,82,145,230
*पिछली प्रश्नोत्तरी का उत्तर*34
शेष सभी संख्याओं में 1 जोड़ने पर पूर्ण वर्ग बन जाती है परन्तु 34 नहीं बनती है।

*Previous Quiz Solution*34
Adding 1 to all the remaining numbers makes a perfect square but does not form 34.
*"This article has been prepared by **Satyam Coaching Centre** on the **Satyam Mathematics** blog."*