250+ TOP MCQs on Polar Curves and Answers

Differential Calculus Multiple Choice Questions & Answers focuses on “Polar Curves”.

1. Polar equations of the circle for the given coordinate (x,y) which satisfies the equation given by (x-a)²+(y-b)²=r² where (a,b) is the coordinates of the centre of the circle &r is the radius.
a) x = r cos⁡θ, y = r sin⁡θ
b) x = a+ r cos⁡θ, y = b + r sin⁡θ
c) y = a+r cos⁡θ, x = b + r sin⁡θ
d) x = r sin⁡θ, y = r cos⁡θ
View Answer

Answer: b
Explanation: option x = a+ r cos⁡θ, y = b + r sin⁡θ satisfies the equation (x-a)²+(y-b)²=r²
because LHS=(a+r cos⁡θ-a)² + (b+ r sin⁡θ-b)² = r²(cos² θ + sin² θ)= r²=RHS.

2. In an polar curve r=f (θ) what is the relation between θ & the coordinates (x,y)?
a) tan⁡θ = (frac{x}{y} )
b) (1+sin⁡θ) = (frac{y}{x} )
c) (1+sec² θ) = (frac{y^2}{x^2} )
d) (1+cos⁡θ) = (frac{x}{y} )
View Answer

Answer: c
Explanation: w.k.t for the polar curve r=f (θ) x=rcos⁡θ, y=rsin⁡θ
dividing them we get (frac{y}{x} = frac{r sin⁡θ}{r cos⁡θ} = tan⁡θ)
squaring on both side (frac{y^2}{x^2} = tan^2 θ = (1+sec^2 θ)).

3. The angle between Radius vector r=a(1-cos⁡θ)and tangent to the curve is ∅ given by _______
a) ∅=(frac{π}{2})
b) ∅=π
c) ∅=(-frac{π}{2})
d) ∅=0
View Answer

Answer: a
Explanation: r= a(1-cos⁡θ)
taking logarithms on both sides we get,
log⁡r = log⁡a + log⁡(1-cos⁡θ)
differentiating w.r.t θ we get,
( frac{1}{r} frac{dr}{dθ} = 0 + frac{sin⁡θ}{1-cos⁡θ})
(frac{1}{r} frac{dr}{dθ} = frac{2 sin ⁡frac{θ}{2} cos⁡frac{θ}{2}}{2sin^2 frac{θ}{2}} = cot⁡frac{θ}{2}) ..(1),
but (cot⁡∅ = frac{1}{r} frac{dr}{dθ})….(2)
From (1)&(2)
∅=(frac{π}{2}).

4. Angle of intersection between two polar curves given by r=a(1+sin⁡θ) & r=a(1-sin⁡θ) is given by ________
a) (frac{π}{4})
b) (frac{π}{2})
c) Π
d) 0
View Answer

Answer: b
Explanation: r=a(1+sin⁡θ) : r=a(1-sin⁡θ)
taking logarithm on both the equations
log⁡r = log⁡a + log⁡(1+sin⁡θ) : log⁡r = log⁡a + log⁡(1-sin⁡θ)
differentiating on both side we get
( frac{1}{r} frac{dr}{dθ} = frac{cos⁡θ}{1+sin⁡θ} : frac{1}{r} frac{dr}{dθ} = frac{-cos⁡θ}{1-sin⁡θ})
(cot⁡∅1 = frac{cos⁡θ}{1+sin⁡θ} : cot⁡∅2 = frac{-cos⁡θ}{1-sin⁡θ})
where ∅1&∅2 are the angle between tangent & the vector respectively
(tan⁡∅1 = frac{1+sin⁡θ}{cos⁡θ} : tan⁡∅2 = frac{1-sin⁡θ}{cos⁡θ})
(tan⁡∅1 . tan⁡∅2 = frac{1+sin⁡θ}{cos⁡θ} . frac{1-sin⁡θ}{-cos⁡θ} = frac{1-sin^2 θ}{-cos^2 θ} = -frac{cos^2 θ}{cos^2 θ} = -1)
above is the condition of orthogonality of two polar curves thus
|∅1-∅2|=(frac{π}{2}).

5. One among the following is the correct explanation of pedal equation of an polar curve, r=f (θ), p=r sin(∅) (where p is the length of the perpendicular from the pole to the tangent & ∅ is the angle made by tangent to the curve with vector drawn to curve from pole)is _______
a) It is expressed in terms of p & θ only
b) It is expressed in terms of p & ∅ only
c) It is expressed in terms of r & θ only
d) It is expressed in terms of p& r only
View Answer

Answer: d
Explanation: It is expressed in terms of p& r only
where p=r sin(∅) & (tan⁡∅ = frac{r}{frac{dr}{dθ}} = r (frac{dr}{dθ}))
& r=f (θ) or after solving we get direct relationship between p & r as
(frac{1}{p^2} = frac{1}{r^2} cosec^2∅.)

6. The pedal Equation of the polar curve rⁿ=aⁿ cos⁡nθ is given by ______
a) rⁿ=paⁿ
b) r^n-1=paⁿ
c) rⁿ⁺¹=paⁿ⁺¹
d) rⁿ⁺¹=paⁿ
View Answer

Answer: d
Explanation: Taking logarithm for the given curve we get
n log⁡r = n log⁡a + log⁡(cos⁡nθ)
differentiating w.r.t θ, we get
(frac{n}{r} frac{dr}{dθ} = frac{-n sin⁡nθ}{cos⁡nθ} rightarrow frac{1}{r} frac{dr}{dθ} = -tan⁡θ)
thus (cot⁡∅ = cot⁡(frac{π}{2} + nθ)rightarrow ∅ = frac{π}{2} + nθ……(1))
from the eqn w.k.t p=r sin ∅
substituting from (1)
p = r sin ((frac{π}{2}) + nθ) = r cos (nθ), but we have rⁿ = aⁿ cos⁡nθ
hence dividing them we get (frac{p}{r^n} = frac{r cos (nθ)}{a^n cos⁡nθ})
rⁿ⁺¹=paⁿ.

7. The length of the perpendicular from the pole to the tangent at the point θ=(frac{π}{2}) on the curve. r=a sec²((frac{π}{2})) is _____
a) (p = frac{2a}{sqrt{3}})
b) (p = frac{4a}{sqrt{3}})
c) (p = 2asqrt{2})
d) p = 4a
View Answer

Answer: c
Explanation: Taking Logarithm on both side of the polar curve
we get log⁡r = log⁡a + 2 log⁡sec⁡((frac{θ}{2}))
differentiating w.r.t θ we get
(frac{1}{r} frac{dr}{dθ} = frac{2 sec⁡(frac{θ}{2}).tan⁡(frac{θ}{2}) }{2 sec⁡(frac{θ}{2})} = tan⁡(frac{θ}{2}))
(cot⁡∅ = cot⁡(frac{π}{2} – frac{θ}{2}) rightarrow ∅ = frac{π}{2} -frac{θ}{2})
w.k.t length of the perpendicular is given by p=r sin ∅
thus substituting ∅ value we get (p = r ,sin(frac{π}{2} – frac{θ}{2}) = r ,cos(frac{θ}{2}))
at ( θ=frac{π}{4}, p = r ,cos frac{π}{4} = frac{r}{sqrt{2}}….(1))
but ( r=a ,sec^2 (frac{θ}{2}) ,at, θ = frac{π}{4}, r=a ,sec^2 (frac{π}{4}) = 4a…(2))
from (1) & (2) (p=frac{4a}{sqrt{2}} = 2asqrt{2}).