10 Most FAQ’s of Tangent and Normal Chapter in Inter 1st Year Maths-1B (TS/AP)

7 Marks

LAQ-1 : If the tangent at a point on the curve x^2/3 + y^2/3 = a^2/3 intersects the coordinate axes in A,B then show that the length AB is a constant

$$P(a\cos 3\theta, a\sin 3\theta) \text{ where } x = a\cos 3\theta \text{ and } y = a\sin 3\theta$$

$$\frac{dx}{d\theta} = -3a\sin 3\theta \quad \text{and} \quad \frac{dy}{d\theta} = 3a\cos 3\theta$$

$$\frac{dy}{dx} = \frac{\frac{dy}{d\theta}}{\frac{dx}{d\theta}} = \frac{3a\cos 3\theta}{-3a\sin 3\theta} = -\frac{\cos \theta}{\sin \theta}$$

$$y – a\sin 3\theta = -\frac{\sin \theta}{\cos \theta}(x – a\cos 3\theta)$$

$$\frac{y – a\sin 3\theta}{\sin \theta} = -\frac{x – a\cos 3\theta}{\cos \theta} \quad \Rightarrow \quad \frac{y}{\sin \theta} – \frac{a\sin 3\theta}{\sin \theta} = -\frac{x}{\cos \theta} + \frac{a\cos 3\theta}{\cos \theta}$$

$$\frac{x}{\cos \theta} + \frac{y}{\sin \theta} = a\cos^2 \theta + a\sin^2 \theta = a(1) \quad \Rightarrow \quad \frac{x}{a\cos \theta} + \frac{y}{a\sin \theta} = 1$$

$$A = (a\cos \theta, 0), \quad B = (0, a\sin \theta)$$

$$AB = \sqrt{(a\cos \theta – 0)^2 + (0 – a\sin \theta)^2} = \sqrt{a^2\cos^2 \theta + a^2\sin^2 \theta} = \sqrt{a^2(\cos^2 \theta + \sin^2 \theta)} = \sqrt{a^2(1)} = a$$

---

LAQ-2 : Show that the tangent at P(x₁,y₁) on the curve √x + √y = √a is xx₁ ^-1/2 + yy₁ ^-1/2 = a^1/2

$$\sqrt{x} + \sqrt{y} = \sqrt{a}$$

$$\frac{1}{2\sqrt{x}} + \frac{1}{2\sqrt{y}} \frac{dy}{dx} = 0$$

$$\frac{1}{2} \left(\frac{1}{\sqrt{x}} + \frac{1}{\sqrt{y}} \frac{dy}{dx}\right) = 0$$

$$\frac{1}{\sqrt{y}} \frac{dy}{dx} = -\frac{1}{\sqrt{x}}$$

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

$$m = -\frac{\sqrt{y_1}}{\sqrt{x_1}}$$

$$y – y_1 = m(x – x_1)$$

$$y – y_1 = -\frac{\sqrt{y_1}}{\sqrt{x_1}} (x – x_1)$$

$$\frac{y – y_1}{\sqrt{y_1}} = -\frac{x – x_1}{\sqrt{x_1}}$$

$$\frac{y}{\sqrt{y_1}} – \frac{y_1}{\sqrt{y_1}} = -\frac{x}{\sqrt{x_1}} + \frac{x_1}{\sqrt{x_1}}$$

$$\frac{x}{\sqrt{x_1}} + \frac{y}{\sqrt{y_1}} = \frac{x_1}{\sqrt{x_1}} + \frac{y_1}{\sqrt{y_1}}$$

$$x x_1^{-1/2} + y y_1^{-1/2} = \sqrt{a}$$

$$x x_1^{-1/2} + y y_1^{-1/2} = a^{1/2}$$

---

LAQ-3 : Find the length of subtangent, subnormal at a point t on the curve x = a(cost + tsint), y = a(sint – tcost)

$$x = a(\cos t + t \sin t)$$

$$y = a(\sin t – t \cos t)$$

$$\frac{dx}{dt} = a \frac{d}{dt}[\cos t + t \sin t] = a[-\sin t + t \cos t + \sin t]$$

$$\frac{dx}{dx} = a(t \cos t)$$

$$\frac{dy}{dt} = a \frac{d}{dt}[\sin t – t \cos t] = a[\cos t – t(-\sin t) + \cos t]$$

$$\frac{dy}{dt} = a(t \sin t)$$

$$m = \frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{a(t \sin t)}{a(t \cos t)} = \tan t$$

$$m = \tan t$$

$$y = a(\sin t – t \cos t) \quad \text{and} \quad m = \tan t$$

---

LAQ-4 : At any point t on the curve x = a(t + sint), y = a(1 – cost) find the lengths of tangent, normal, subtangent and subnormal

$$x = a(t + \sin t)$$

$$\frac{dx}{dt} = a \frac{d}{dt}(t + \sin t) = a(1 + \cos t)$$

Also given:

$$y = a(1 – \cos t)$$

$$\frac{dy}{dt} = a \frac{d}{dt}(1 – \cos t) = a(0 + \sin t) = a \sin t$$

$$m = \frac{dy}{dx} = \frac{a(\sin t)}{a(1 + \cos t)} = \frac{\sin t}{1 + \cos t} = \tan \frac{t}{2}$$

(i) Length of tangent:

$$|y \sqrt{1 + m^2} / m| = |a(1 – \cos t) \sqrt{1 + \tan^2 \frac{t}{2}} / \tan \frac{t}{2}| = |2a \sin \frac{t}{2}|$$

(ii) Length of normal:

$$|y \sqrt{1 + m^2}| = |a(1 – \cos t) \sqrt{1 + \tan^2 \frac{t}{2}}| = |2a \sin^2 \frac{t}{2} / \cos \frac{t}{2}| = |2a \sin t / 2 \tan t / 2|$$

(iii) Length of subtangent:

$$|y / m| = |a(1 – \cos t) / \tan \frac{t}{2}| = |2a \sin^2 \frac{t}{2} \cos \frac{t}{2} / \sin \frac{t}{2}| = |a \sin t|$$

(iv) Length of subnormal:

$$|y \cdot m| = |a(1 – \cos t) \tan \frac{t}{2}| = |2a \sin^2 \frac{t}{2} \tan \frac{t}{2}|$$

---

LAQ-5 : Find the angle between the curves xy = 2 and x² + 4y = 0

1. Finding the Point of Intersection:

$$xy = 2 \quad \text{(1)}$$

$$x^2 + 4y = 0 \quad \text{(2)}$$

Substitute (1) into (2) and solve for xxx:

$$x^2 + 4\left(\frac{2}{x}\right) = 0$$

$$x^2 + \frac{8}{x} = 0$$

$$x^3 + 8 = 0$$

$$x^3 = -8$$

$$x = -2$$

$$y = \frac{2}{x} = \frac{2}{-2} = -1$$

2. Finding Derivatives:

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

$$\frac{dy}{dx} = -\frac{y}{x} \quad \text{(3)}$$

$$2x + 4 \frac{dy}{dx} = 0$$

$$\frac{dy}{dx} = -\frac{x}{2} \quad \text{(4)}$$

3. Finding Slopes at P(−2,−1):

$$m_1 = -\frac{y}{x} = -\frac{-1}{-2} = -\frac{1}{2} \quad \text{(5)}$$

$$m_2 = -\frac{x}{2} = -\frac{-2}{2} = 1 \quad \text{(6)}$$

4. Finding the Angle θ at Point P:

$$\tan \theta = \left|\frac{m_1 – m_2}{1 + m_1 m_2}\right| = \left|\frac{-\frac{1}{2} – 1}{1 + (-\frac{1}{2} \cdot 1)}\right| = \left|\frac{-\frac{3}{2}}{\frac{1}{2}}\right| = 3$$

$$\theta = \tan^{-1}(3)$$

---

LAQ-6 : Find the angle between the curves 2y² – 9x = 0, 3x² + 4y = 0 (in Q₄)

1) Finding the Point of Intersection:

$$2y^2 – 9x = 0 \quad \text{(1)}$$

$$3x^2 + 4y = 0 \quad \text{(2)}$$

Substitute equation (1) into equation (2):

$$x = \frac{2}{9}y^2 \quad \text{(from equation (1))}$$

$$3\left(\frac{2}{9}y^2\right)^2 + 4y = 0$$

$$\frac{4}{27}y^4 + 4y = 0$$

$$4y\left(\frac{y^3}{27} + 1\right) = 0$$

$$y(y^3 + 27) = 0$$

$$y = 0 \quad \text{or} \quad y^3 = -27$$

$$y = -3, \quad x = \frac{2}{9} \times 9 = 2$$

$$P(x, y) = (2, -3)$$

2) Finding Derivatives:

$$4y \frac{dy}{dx} – 9 = 0$$

$$\frac{dy}{dx} = \frac{9}{4y} \quad \text{(3)}$$

$$6x + 4 \frac{dy}{dx} = 0$$

$$\frac{dy}{dx} = -\frac{3x}{2} \quad \text{(4)}$$

3) Finding Slopes at P(2,−3):

$$m_1 = \frac{9}{4(-3)} = -\frac{3}{4} \quad \text{(5)}$$

$$m_2 = -\frac{3 \times 2}{2} = -3 \quad \text{(6)}$$

4) Finding the Angle at P:

$$\tan \theta = \left| \frac{m_1 – m_2}{1 + m_1 m_2} \right| = \left| \frac{-\frac{3}{4} + 3}{1 + (-\frac{3}{4} \cdot -3)} \right| = \left| \frac{-3 + 12}{4}/ \frac{9 + 12}{4} \right| = \frac{9}{13}$$

$$\theta = \tan^{-1}\left(\frac{9}{13}\right)$$

---

LAQ-7 : Find the angle between the curves y² = 4x and x² + y² = 5

1) Finding Points of Intersection:

$$y^2 = 4x \quad \text{(1)}$$

$$x^2 + y^2 = 5 \quad \text{(2)}$$

Combine (1) & (2) to solve for x and y:

$$x^2 + 4x = 5$$

$$x^2 + 4x – 5 = 0$$

$$(x – 1)(x + 5) = 0$$

$$x = 1 \quad \text{or} \quad x = -5$$

$$\text{If } x = 1, \quad y^2 = 4 \times 1 = 4 \quad \Rightarrow \quad y = \pm 2$$

$$P = (1, 2), \quad Q = (1, -2)$$

2) Finding Derivatives:

$$2y \frac{dy}{dx} = 4$$

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

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

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

3) Finding Slopes at P(1,2) and Q(1,−2):

At P(1,2):

$$m_1 = \frac{2}{2} = 1;$$

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

At Q(1,−2):

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

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

4) Finding the Angle at P and Q:

$$\tan \theta = \left|\frac{m_1 – m_2}{1 + m_1 m_2}\right| = \left|\frac{1 – (-\frac{1}{2})}{1 + 1(-\frac{1}{2})}\right| = \frac{3}{2} / \frac{1}{2} = 3$$

$$\theta = \tan^{-1}(3)$$

$$\tan \theta = \left|\frac{m_1 – m_2}{1 + m_1 m_2}\right| = \left|\frac{-1 – \frac{1}{2}}{1 + (-1)(\frac{1}{2})}\right| = \frac{-3}{2} / \frac{1}{2} = -3$$

$$\theta = \tan^{-1}(-3)$$

---

LAQ-8 : Find the angle between the curves y² = 8x and 4x² + y² = 32

1) Finding Points of Intersection:

$$y^2 = 8x \quad \text{(1)}$$

$$4x^2 + y^2 = 32 \quad \text{(2)}$$

Combine (1) & (2) and solve for x and y:

$$4x^2 + 8x = 32$$

$$4(x^2 + 2x – 8) = 0$$

$$x^2 + 2x – 8 = 0$$

$$(x + 4)(x – 2) = 0$$

$$x = -4, \quad x = 2$$

$$\text{If } x = 2, \quad y^2 = 8 \times 2 = 16 \quad \Rightarrow \quad y = \pm 4$$

$$P = (2, 4), \quad Q = (2, -4)$$

2) Finding Derivatives:

$$2y \frac{dy}{dx} = 8$$

$$\frac{dy}{dx} = \frac{4}{y}$$

$$8x + 2y \frac{dy}{dx} = 0$$

$$2y \frac{dy}{dx} = -8x$$

$$\frac{dy}{dx} = -\frac{4x}{y}$$

3) Finding Slopes at P(2,4):

$$m_1 = \frac{4}{4} = 1$$

$$m_2 = -\frac{4 \times 2}{4} = -2$$

4) Finding the Angle at P:

$$\tan \theta = \left| \frac{m_1 – m_2}{1 + m_1 m_2} \right| = \left| \frac{1 + 2}{1 – 2} \right| = 3$$

$$\theta = \tan^{-1}(3)$$

5) Finding Slopes at Q(2,−4):

$$m_1 = \frac{4}{-4} = -1$$

$$m_2 = -\frac{4 \times 2}{-4} = 2$$

6) Finding the Angle at Q:

$$\tan \theta = \left| \frac{-1 – 2}{1 + (-1) \times 2} \right| = \left| \frac{-3}{-1} \right| = 3$$

$$\theta = \tan^{-1}(3)$$

---

LAQ-9 : Show that the curves y² = 4(x + 1), y² = 36(9 – x) intersect orthogonally

1) Finding Points of Intersection:

$$y^2 = 4(x + 1) \quad \text{(1)}$$

$$y^2 = 36(9 – x) \quad \text{(2)}$$

Equating and solving (1) & (2):

$$4(x + 1) = 36(9 – x)$$

$$x + 1 = 9(9 – x)$$

$$x + 1 = 81 – 9x$$

$$10x = 80$$

$$x = 8$$

$$y^2 = 4(8 + 1) = 36$$

$$y = \pm 6$$

2) Finding Derivatives:

$$2y \frac{dy}{dx} = 4$$

$$\frac{dy}{dx} = \frac{2}{y} \quad \text{(3)}$$

$$2y \frac{dy}{dx} = -36$$

$$\frac{dy}{dx} = -\frac{18}{y} \quad \text{(4)}$$

3) Finding Slopes at Points P(8,6) and Q(8,−6):

$$m_1 = \frac{2}{6} = \frac{1}{3}$$

$$m_2 = -\frac{18}{6} = -3$$

$$m_1 \cdot m_2 = \frac{1}{3} \cdot (-3) = -1$$

$$m_1 = \frac{2}{-6} = -\frac{1}{3}$$

$$m_2 = -\frac{18}{-6} = 3$$

---

LAQ-10 : Find the condition for the orthogonally of the curves ax² + by² = 1 and a₁x² + b₁y² = 1

1) Finding Point of Intersection:

$$ax^2 + by^2 = 1 \quad \text{(Equation 1)}$$

$$a_1x^2 + b_1y^2 = 1 \quad \text{(Equation 2)}$$

$$ax_1^2 + by_1^2 = a_1x_1^2 + b_1y_1^2$$

$$ax_1^2 – a_1x_1^2 = b_1y_1^2 – by_1^2$$

$$x_1^2(a – a_1) = y_1^2(b_1 – b)$$

$$\frac{x_1^2}{y_1^2} = \frac{b_1 – b}{a – a_1} = -\frac{b – b_1}{a – a_1} \quad \text{(1)}$$

2) Finding Derivatives & Slopes:

$$2ax + 2by \frac{dy}{dx} = 0$$

$$\frac{dy}{dx} = -\frac{ax}{by} \quad \text{(Slope of the tangent at ( P(x_1, y_1) ))}$$

$$2a_1x + 2b_1y \frac{dy}{dx} = 0$$

$$\frac{dy}{dx} = -\frac{a_1x}{b_1y} \quad \text{(Slope of the tangent at ( P(x_1, y_1) ))}$$

3) Applying Orthogonal Condition:

$$\left(-\frac{ax_1}{by_1}\right)\left(-\frac{a_1x_1}{b_1y_1}\right) = -1$$

$$\frac{a a_1 x_1^2}{b b_1 y_1^2} = -1$$

$$\frac{x_1^2}{y_1^2} = -\frac{b b_1}{a a_1} \quad \text{(2)}$$

Equating Equations (1) and (2):

$$\frac{b b_1}{a a_1} = \frac{b – b_1}{a – a_1}$$

$$\frac{a – a_1}{a a_1} = \frac{b – b_1}{b b_1}$$

$$\frac{1}{a_1} – \frac{1}{a} = \frac{1}{b_1} – \frac{1}{b}$$