250+ TOP MCQs on Orthogonal Trajectories and Answers

Ordinary Differential Equations Multiple Choice Questions on “Orthogonal Trajectories”.

1. Find the orthogonal trajectories of the family of parabolas y²=4ax.
a) 2x²+y²=k
b) 2y²+x²=k
c) x²-2y²=k
d) 2x²-y²=k
Answer: a
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 y²=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 2x²+y²=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) x²-y²-3a² log⁡x-k = 0
b) x²+2y²–(frac{a^2}{2}) log⁡x-k = 0
c) x²+y²-2a² log⁡x-k = 0
d) 2x²-y²–(frac{a^2}{3}) log⁡x-k = 0
Answer: c
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)
–> x²+y²-2a² log⁡x – 2c=0 or x²+y²-2a² log⁡x-k=0 where k=2c is the required orthogonal trajectory.

3. The Orthogonal DE for family of parabola y²=4a(x+a) is same as _______(where DE stands for Differential equation)
a) DE of parabola y²=4a(x+a)
b) DE of parabola y²=4ax
c) DE of parabola x²=4ay
d) DE of parabola x²=4a(y+a)
Answer: a
Explanation: y²=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=2xfrac{-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) r²=k(cos θ)²
c) r=k(1-cos θ)
d) r=k(1-sin θ)
Answer: d
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{cos⁡θ}{1+sin⁡θ}) since given equation is polar slope of tangent is given by ( frac{1}{r} frac{dr}{dθ}) and perpendicular line has a slope of (-rfrac{dθ}{dr} , replacing, frac{1}{r} frac{dr}{dθ} ,by, -rfrac{dθ}{dr})
we have (-rfrac{dθ}{dr} = frac{cos⁡θ}{1+sin⁡θ}) …..separating the variables and integrating it
(int frac{dr}{r} + int frac{1+sin⁡θ}{cos⁡θ} ,dθ = c rightarrow log r + int sec⁡θ ,dθ + int tan⁡θ , dθ = c)
log r + log(sec⁡θ + tan⁡θ) + log(sec⁡θ) = c = log k
(rightarrow log(r(sec⁡θ + tan⁡θ) (sec⁡θ)) = log k rightarrow rleft(frac{1}{cos⁡θ} + frac{sin⁡θ}{cos⁡θ}right) frac{1}{cos⁡θ} = k)
(frac{r(1+sin⁡θ)}{cos^2 θ} = frac{r(1+sin⁡θ)}{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 rⁿ = a sin⁡nθ?
a) Given family of curve is Self orthogonal
b) Orthogonal trajectory is rⁿ=k cos⁡nθ where k is an constant
c) Orthogonal trajectory is rⁿ=k cosec⁡nθ where k is an constant
d) Orthogonal trajectory is rⁿ=k sin⁡nθ where k is an constant
Answer: b
Explanation: Consider rⁿ = a sin⁡nθ..(1) –> n log r = log a + log(sin⁡nθ) …differentiating w.r.t θ
( frac{n}{r} frac{dr}{dθ} = frac{n cos⁡nθ}{sin⁡nθ} ,or, frac{1}{r} frac{dr}{dθ} = frac{cos⁡nθ}{sin⁡nθ} = cot ,nθ)
replacing (frac{1}{r} frac{dr}{dθ} ,by, -rfrac{dθ}{dr} ,we,, get, -rfrac{dθ}{dr} = cot ,nθ)
separating the variables and integrating (int frac{dr}{r} int tan⁡,nθ ,dθ = c )
( rightarrow log ,r + frac{log⁡(sec⁡nθ)}{n} = c) or n log r + log⁡(sec nθ) = nc = log⁡k
log(rⁿ sec⁡nθ) = log k –> rⁿ sec⁡ nθ=k or rⁿ=k cos⁡nθ ..(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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