This set of Ordinary Differential Equations Multiple Choice Questions & Answers (MCQs) focuses on “Orthogonal Trajectories”.

1. Find the orthogonal trajectories of the family of parabolas y^{2}=4ax.

a) 2x^{2}+y^{2}=k

b) 2y^{2}+x^{2}=k

c) x^{2}-2y^{2}=k

d) 2x^{2}-y^{2}=k

View Answer

Explanation: Consider \(\frac{y^2}{x} = 4a\)……..differentiating w.r.t x we get \(\frac{x.2y \frac{dy}{dx} – y^2.1}{x^2} = 0\)

\(2xy \frac{dy}{dx} -y = 0\)…..DE of the family of curve y

^{2}=4ax replacing \(\frac{dy}{dx}\) by \(-\frac{dx}{dy}\)

since \(\frac{dy}{dx}\) is the slope of the tangent to the curve –> \(-\frac{dx}{dy}\) is the slope of orthogonal line we get \(2x(-\frac{dx}{dy}) – y = 0 \, or\, 2x \,dx + y \,dy = 0 \rightarrow \int2x \,dx + \int y \,dy = c \rightarrow x^2 + \frac{y^2}{2} = c.\) or 2x

^{2}+y

^{2}=k is the required orthogonal trajectory.

2. Find the orthogonal trajectories of the family of curves \(\frac{x^2}{a^2} + \frac{y^2}{b^2+k} = 1\) where k is the parameter.

a) x^{2}-y^{2}-3a^{2} logx-k = 0

b) x^{2}+2y^{2}–\(\frac{a^2}{2}\) logx-k = 0

c) x^{2}+y^{2}-2a^{2} logx-k = 0

d) 2x^{2}-y^{2}–\(\frac{a^2}{3}\) logx-k = 0

View Answer

Explanation: \(\frac{x^2}{a^2} + \frac{y^2}{b^2+k} = 1\)…….(1) differentiating w.r.t x we have \(\frac{2x}{a^2} + \frac{2yy’}{b^2+k} = 0 \,where\, y’ = \frac{dy}{dx}\)

i.e \(\frac{x^2}{a^2} = \frac{-yy’}{b^2+k} …..(2) \,from\, (1)\, \frac{x^2}{a^2} – 1 = \frac{-y^2}{b^2+k} \rightarrow \frac{x^2-a^2}{a^2} = \frac{-y^2}{b^2+k}\)…..(3) divide (2)&(3)

we get \(\frac{x}{x^2-a^2} = \frac{yy’}{y^2} \rightarrow \frac{x}{x^2-a^2} = \frac{y’}{y}\)

now \(y’=\frac{dy}{dx}\) is replaced by \(-\frac{dx}{dy} \,i.e\, \frac{x}{x^2-a^2} = \frac{1}{y}(-\frac{dx}{dy})\) separating the variable we get \(y \,dy = -\frac{(x^2-a^2)}{x} dx\) integrating this equation

\(\int y \,dy = \int-x \,dx + \int a^2 \frac{1}{x} \,dx + c \rightarrow \frac{y^2}{2} = \frac{-x^2}{2} + a^2 log \,x + c\)

–> x

^{2}+y

^{2}-2a

^{2}logx – 2c=0 or x

^{2}+y

^{2}-2a

^{2}logx-k=0 where k=2c is the required orthogonal trajectory.

3. The Orthogonal DE for family of parabola y^{2}=4a(x+a) is same as _______(where DE stands for Differential equation)

a) DE of parabola y^{2}=4a(x+a)

b) DE of parabola y^{2}=4ax

c) DE of parabola x^{2}=4ay

d) DE of parabola x^{2}=4a(y+a)

View Answer

Explanation: y

^{2}=4a(x+a)……….differentiating w.r.t x we get \(2y \frac{dy}{dx} = 4a \rightarrow a =\frac{yy’}{2}\)

substituting the ‘a’ in given equation i.e : \(y^2 = 2yy’(x + \frac{yy’}{2}), y=2xy’+yy’^2\)…(1)

replacing y’ by \(\frac{-1}{y’} \,i.e\, y=2x\frac{-1}{y’} + y(\frac{-1}{y’})^2\) or on solving yy’

^{2}+2xy’=y…(2)

comparing (2) & (1) we observe that DE of the orthogonal family is same as the DE of the given family of curves thus the family of parabola is self orthogonal.

4. Find the orthogonal trajectories of the family r=a(1+sin θ).

a) r=k(sin θ)

b) r^{2}=k(cos θ)^{2}

c) r=k(1-cos θ)

d) r=k(1-sin θ)

View Answer

Explanation: r=a(1+sin θ) taking log we get log r=log a+log(1+sin θ)…differentiating

w.r.t θ we have, \( \frac{1}{r} \frac{dr}{dθ} = \frac{cosθ}{1+sinθ}\) since given equation is polar slope of tangent is given by \( \frac{1}{r} \frac{dr}{dθ}\) and perpendicular line has a slope of \(-r\frac{dθ}{dr} \, replacing\, \frac{1}{r} \frac{dr}{dθ} \,by\, -r\frac{dθ}{dr}\)

we have \(-r\frac{dθ}{dr} = \frac{cosθ}{1+sinθ}\) …..separating the variables and integrating it

\(\int \frac{dr}{r} + \int \frac{1+sinθ}{cosθ} \,dθ = c \rightarrow log r + \int secθ \,dθ + \int tanθ \, dθ = c\)

log r + log(secθ + tanθ) + log(secθ) = c = log k

\(\rightarrow log(r(secθ + tanθ) (secθ)) = log k \rightarrow r\left(\frac{1}{cosθ} + \frac{sinθ}{cosθ}\right) \frac{1}{cosθ} = k\)

\(\frac{r(1+sinθ)}{cos^2 θ} = \frac{r(1+sinθ)}{1-sin^2 θ} \rightarrow r = k(1-sin θ)\) is the required orthogonal trajectory.

5. Which among the following is true for the curve r^{n} = a sinnθ?

a) Given family of curve is Self orthogonal

b) Orthogonal trajectory is r^{n}=k cosnθ where k is an constant

c) Orthogonal trajectory is r^{n}=k cosecnθ where k is an constant

d) Orthogonal trajectory is r^{n}=k sinnθ where k is an constant

View Answer

Explanation: Consider r

^{n}= a sinnθ..(1) –> n log r = log a + log(sinnθ) …differentiating w.r.t θ

\( \frac{n}{r} \frac{dr}{dθ} = \frac{n cosnθ}{sinnθ} \,or\, \frac{1}{r} \frac{dr}{dθ} = \frac{cosnθ}{sinnθ} = cot \,nθ\)

replacing \(\frac{1}{r} \frac{dr}{dθ} \,by\, -r\frac{dθ}{dr} \,we\,\, get\, -r\frac{dθ}{dr} = cot \,nθ\)

separating the variables and integrating \(\int \frac{dr}{r} \int tan\,nθ \,dθ = c \)

\( \rightarrow log \,r + \frac{log(secnθ)}{n} = c\) or n log r + log(sec nθ) = nc = logk

log(r

^{n}secnθ) = log k –> r

^{n}sec nθ=k or r

^{n}=k cosnθ ..(2) is the required orthogonal trajectory since (1)&(2) are not same the given family of curve is not self orthogonal.

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