This set of Differential and Integral Calculus Quiz focuses on “Envelopes”.

1. Envelope of a family of curves can be defined as __________

a) A curve which touches 50% of the family of curves

b) A curve which is a straight line

c) A curve which touches each member of the family of curves

d) A curve which surrounds the family of curves

View Answer

Explanation: An envelope is a curve that is a tangent to all the members of the family of curves at some point. It intersects all the curves at some point.

2. Do concentric circles have an envelope?

a) Yes

b) No

View Answer

Explanation: Concentric circles are circles having the same center but different radii. No curve can be a tangent to all the circles. Hence, there is no envelope for concentric circles.

3. What is the relation between evolutes and envelopes?

a) Evolutes and envelopes are same

b) Evolute is the envelope of normals to a curve

c) Evolute is the envelope of tangents to a curve

d) Envelope is the evolute of normal to a curve

View Answer

Explanation: Evolute is the locus of the all the centres of curvature of the curve whereas envelope is the curve which touches all the members of the family of the curve i.e Envelope is tangent to all the curves in a family of curves. Hence, evolute is the envelope of normal to a curve.

4. Find the envelope of the family of lines \(\frac{x}{t} \) + yt = 2c, t being the parameter.

a) xy = c

b) xy = 2c

c) xy = 2

d) xy = c^{2}

View Answer

Explanation: Given equation can be written as yt

^{2}– 2ct + x = 0 ——–> eq(1)

The envelope of At

^{2}+ Bt + C = 0 is B

^{2}– 4AC = 0 ———> eq(2)

From eq(1), A = y, B= -2c, C = x

Putting the values in eq(2),

(-2c)

^{2}– 4(y)(x) = 0

4c

^{2}– 4xy = 0

xy = c

^{2}.

5. What is the envelope of the family of straight lines y = mx +\(\frac{a}{m} \), m is the parameter?

a) y = 4ax

b) y^{2} = 4ax

c) x = 4ay

d) x^{2} = 4ay

View Answer

Explanation: The given equation can be written as m

^{2}x – ym + a = 0 —-> eq(1) is in the form At

^{2}+ Bt + C =0

The envelope is given by B

^{2}– 4AC = 0 ——> eq(2)

From eq(1), A = x , B = -y , C = a

Putting the values in eq(2),

(-y)

^{2}– 4(x)(a) = 0

y

^{2}= 4ax.

6. What is the envelope of straight lines given by x cos b + y sin b = a sec b, where b is the parameter?

a) y^{2} + 4a(a-x) = 0

b) y + 4ax = 0

c) y + 4a(a-x) = 0

d) y^{2} = 4a(a-x)

View Answer

Explanation: Dividing the given equation by cos b,

x + y tan b = a \(\frac{secb}{cosb}\) = a sec

^{2}b = a(1 + tan

^{2}b)

The above equation can be written as a tan

^{2}b – y tan b + (a-x) = 0 which is a quadratic equation in tan b

Hence, A = a, B = -y, C = a-x

The envelope is B

^{2}– 4AC = 0

(-y)

^{2}– 4(a)(a-x) = 0

y

^{2}= 4a(a-x).

**Sanfoundry Global Education & Learning Series – Differential and Integral Calculus.**

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