10.2 Tangent and Normal (SAQs)

Maths-1B | 10.2 Tangent and Normal – SAQs:
Welcome to SAQs in Chapter 10: 10.2 Tangent and Normal. This page contains the most Important FAQs for Short Answer Questions in this Chapter. Each answer is provided in simple and easy-to-understand steps. This will support your preparation and help you secure top marks in your exams.

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SAQ-1 : Find the equations of the tangent and the normal to the curve xy = 10 at (2,5)

Given Curve and Slope Calculation:

$$xy = 10 \quad \text{(Equation of the curve)}$$

$$y = \frac{10}{x}$$

$$\frac{dy}{dx} = -\frac{10}{x^2}$$

$$m = \left(\frac{dy}{dx}\right)_{(2,5)} = -\frac{10}{2^2} = -\frac{10}{4} = -\frac{5}{2}$$

1. Equation of the Tangent at (2,5):

$$y – y_1 = m(x – x_1)$$

$$y – 5 = -\frac{5}{2}(x – 2)$$

$$2y – 10 = -5x + 10$$

$$5x + 2y – 20 = 0$$

2. Equation of the Normal:

$$\text{Slope of the normal} = -\frac{1}{m} = \frac{2}{5}$$

$$y – y_1 = \left(-\frac{1}{m}\right)(x – x_1)$$

$$y – 5 = \frac{2}{5}(x – 2)$$

$$5y – 25 = 2x – 4$$

$$2x – 5y + 21 = 0$$

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SAQ-2 : Find the equations of the tangent and the normal to the curve y = x³ + 4x² at (-1,3)

Given Curve and Differentiation:

$$y = x^3 + 4x^2 \quad \text{(Equation of the curve)}$$

$$\frac{dy}{dx} = 3x^2 + 8x$$

$$m = 3(-1)^2 + 8(-1) = 3 – 8 = -5$$

1. Equation of the Tangent at P(−1,3):

$$y – y_1 = m(x – x_1)$$

$$y – 3 = -5(x + 1)$$

$$5x + y + 2 = 0$$

2. Equation of the Normal:

$$\text{Slope of the normal} = -\frac{1}{m} = \frac{1}{5}$$

$$y – y_1 = \left(-\frac{1}{m}\right)(x – x_1)$$

$$y – 3 = \frac{1}{5}(x + 1)$$

$$5y – 15 = x + 1$$

$$x – 5y + 16 = 0$$

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SAQ-3 : Find the equations of the tangent and the normal to the curve y⁴ = ax³ at (a, a)

Given Curve and Differentiation:

$$y^4 = ax^3 \quad \text{(Equation of the curve)}$$

$$4y^3 \frac{dy}{dx} = 3ax^2$$

$$\frac{dy}{dx} = \frac{3ax^2}{4y^3}$$

$$m = \frac{3a \cdot a^2}{4a^3} = \frac{3}{4}$$

1. Equation of the Tangent at (a,a):

$$y – y_1 = m(x – x_1)$$

$$y – a = \frac{3}{4}(x – a)$$

$$4(y – a) = 3(x – a)$$

$$4y – 4a = 3x – 3a$$

$$3x – 4y + a = 0$$

2. Equation of the Normal:

$$\text{Slope of the normal} = -\frac{1}{m} = -\frac{4}{3}$$

$$y – y_1 = \left(-\frac{1}{m}\right)(x – x_1)$$

$$y – a = -\frac{4}{3}(x – a)$$

$$3(y – a) = -4(x – a)$$

$$3y – 3a = -4x + 4a$$

$$4x + 3y – 7a = 0$$

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SAQ-4 : Show that the curves x² + y² = 2, 3x² + y² = 4x have a common tangent at the point (1, 1)

Given First Curve and Differentiation:

$$x^2 + y^2 = 2 \quad \text{(Equation of the first curve)}$$

$$2x + 2y \frac{dy}{dx} = 0$$

$$2y \frac{dy}{dx} = -2x$$

$$\frac{dy}{dx} = -\frac{x}{y}$$

$$m_1 = -\frac{1}{1} = -1 \quad \text{(1)}$$

Given Second Curve and Differentiation:

$$3x^2 + y^2 = 4x \quad \text{(Equation of the second curve)}$$

$$6x + 2y \frac{dy}{dx} = 4$$

$$2y \frac{dy}{dx} = 4 – 6x$$

$$\frac{dy}{dx} = \frac{2 – 3x}{y}$$

$$m_2 = \frac{2 – 3 \times 1}{1} = -1 \quad \text{(2)}$$

$$m_1 = m_2 = -1$$

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SAQ-5 : Show that the curves 6x² – 5x + 2y = 0, 4x² + 8y² = 3 touch each other at (1/2, 1/2)

Given First Curve and Differentiation:

$$6x^2 – 5x + 2y = 0 \quad \text{(Equation of the first curve)}$$

$$12x – 5 + 2 \frac{dy}{dx} = 0$$

$$2 \frac{dy}{dx} = 5 – 12x$$

$$\frac{dy}{dx} = \frac{5 – 12x}{2}$$

$$m_1 = \frac{5 – 12 \cdot \frac{1}{2}}{2} = \frac{5 – 6}{2} = -\frac{1}{2} \quad \text{(1)}$$

Given Second Curve and Differentiation:

$$4x^2 + 8y^2 = 3 \quad \text{(Equation of the second curve)}$$

$$8x + 16y \frac{dy}{dx} = 0$$

$$x + 2y \frac{dy}{dx} = 0$$

$$2y \frac{dy}{dx} = -x$$

$$\frac{dy}{dx} = -\frac{x}{2y}$$

$$m_2 = -\frac{\frac{1}{2}}{2 \cdot \frac{1}{2}} = -\frac{1}{2} \quad \text{(2)}$$

$$m_1 = m_2 = -\frac{1}{2}$$

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SAQ-6 : Find the length of subtangent and subnormal at a point on the curve y = bsin(x/a)

$$y = b \sin\left(\frac{x}{a}\right) \quad \text{(Equation of the curve)}$$

$$\frac{dy}{dx} = b \cos\left(\frac{x}{a}\right)\frac{1}{a}$$

1. Length of Sub-Tangent:

$$\text{Length of sub-tangent} = \left|\frac{y}{m}\right| = \left|\frac{b \sin\left(\frac{x}{a}\right)}{b/a \cos\left(\frac{x}{a}\right)}\right| = \left|a \tan\left(\frac{x}{a}\right)\right|$$

2. Length of Sub-Normal:

$$\text{Length of sub-normal} = \left|y \cdot m\right| = \left|b \sin\left(\frac{x}{a}\right) \cdot b/a \cos\left(\frac{x}{a}\right)\right|$$

$$= \left|\frac{b^2}{a} \frac{1}{2} 2 \sin\left(\frac{x}{a}\right) \cos\left(\frac{x}{a}\right)\right| = \left|\frac{b^2}{2a} \sin\left(\frac{2x}{a}\right)\right|$$

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SAQ-7 : Show that at any point (x,y) on the curve y = be^x/a the length of subtangent is a constant and the length of the subnormal is y²/a

$$y = b e^{\frac{x}{a}} \quad \text{(Equation of the curve)}$$

$$\frac{dy}{dx} = b e^{\frac{x}{a}} \cdot \frac{1}{a} = \frac{b}{a} e^{\frac{x}{a}}$$

1. Length of Subtangent:

$$\text{Length of subtangent} = \left|\frac{y}{m}\right| = \left|\frac{b e^{\frac{x}{a}}}{\frac{b}{a} e^{\frac{x}{a}}}\right| = \left|\frac{b}{b/a}\right| = a$$

2. Length of Subnormal:

$$\text{Length of subnormal} = \left|y \cdot m\right| = \left|b e^{\frac{x}{a}} \cdot \frac{b}{a} e^{\frac{x}{a}}\right| = \left|\frac{b^2}{a} \left(e^{\frac{x}{a}}\right)^2\right| = \left|\frac{b^2}{a} e^{\frac{2x}{a}}\right|$$

$$= \left|\frac{y^2}{a}\right|$$

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SAQ-8 : Find the value of k, so that the length of the subnormal at any point on the curve y = a^1-kx^k is a constant

$$y = a x^{1-k} \quad \text{(Equation of the curve)}$$

$$\frac{dy}{dx} = a(1-k)x^{1-k-1} \quad \text{(where ( m ) is the derivative)}$$

$$\text{Length of subnormal} = \left|y \cdot m\right| = \left|a x^{1-k} \cdot a(1-k) x^{1-k-1}\right|$$

$$= \left|a^2 (1-k) x^{(1-k) + (1-k-1)}\right| = \left|a^2 (1-k) x^{2(1-k) – 1}\right|$$

$$= \left|k a^2 x^{2(1-k) – 1}\right|$$

$$2(1-k) – 1 = 0$$

$$2 – 2k – 1 = 0 \
1 – 2k = 0 \
2k = 1 \
k = \frac{1}{2}$$

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SAQ-9 : S.T the tangent at any point θ on the curve x=c secθ, y=c tanθ is ysinθ = x – ccosθ

$$x = c \sec \theta, \quad y = c \tan \theta$$

$$\frac{dx}{d\theta} = c \sec \theta \tan \theta, \quad \frac{dy}{d\theta} = c \sec^2 \theta$$

$$\frac{dy}{dx} = \frac{\frac{dy}{d\theta}}{\frac{dx}{d\theta}} = \frac{c \sec^2 \theta}{c \sec \theta \tan \theta} = \frac{\sec \theta}{\tan \theta} = \cot \theta$$

$$= \frac{\cos \theta}{\sin \theta}$$

$$\frac{dy}{dx} = \frac{1}{\sin \theta}$$

$$y – y_1 = m(x – x_1)$$

$$y – c \tan \theta = \frac{1}{\sin \theta}(x – c \sec \theta)$$

$$y – \frac{c \sin \theta}{\cos \theta} = \frac{x}{\sin \theta} – \frac{c}{\cos \theta \sin \theta}$$

$$y \sin \theta = x – \frac{c}{\cos \theta} + \frac{c \sin^2 \theta}{\cos \theta}$$

$$y \sin \theta = x – \frac{c \cos^2 \theta}{\cos \theta}$$

$$y \sin \theta = x – c \cos \theta$$

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SAQ-10 : Find the equations of the tangent to the curve y = 3x² – x³, where it meets the X-axis

Finding Points of Intersection with the X-axis:

$$y = 0$$

$$0 = 3x^2 – x^3$$

$$x^2(3-x) = 0$$

$$x = 0, \quad x = 3$$

$$O(0,0) \quad \text{and} \quad A(3,0)$$

$$y = 3x^2 – x^3$$

$$\frac{dy}{dx} = 6x – 3x^2$$

$$m_1 = 6(0) – 3(0)^2 = 0$$

$$y – 0 = 0(x-0)$$

$$y = 0$$

$$m_2 = 6(3) – 3(3)^2 = 18 – 27 = -9$$

$$y – 0 = -9(x-3)$$

$$y = -9x + 27$$

$$9x + y – 27 = 0$$