第 12 类 RD Sharma 解决方案 - 第 16 章切线和法线 - 练习 16.2 |设置 1

问题 1. 求曲线 √x + √y = a 在点 (a ² /4, a ² /4) 的切线方程。

解决方案：

We have,

√x + √y = a

On differentiating both sides w.r.t. x, we get

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

dy/dx = -√y/√x

Given, (x₁, y₁) = (a²/4, a²/4), 

Slope of tangent, m = $(\frac{dy}{dx})_{\left(\frac{a^2}{4},\frac{a^2}{4}\right)}=\frac{-\sqrt{\frac{a^2}{4}}}{\sqrt{\frac{a^2}{4}}}=-1$

The equation of tangent is,

y – y₁ = m (x – x₁)

y – a²/4 = –1(x – a²/4)

y – a²/4 = –x + a²/4

x + y = a²/2

问题 2. 求 y = 2x ³ − x ² + 3 在 (1, 4) 处的法线方程。

解决方案：

We have,

y = 2x³ − x² + 3

On differentiating both sides w.r.t. x, we get

dy/dx = 6x² – 2x

Slope of tangent = $\left( \frac{dy}{dx} \right)_{\left( 1, 4 \right)}$ = 6 (1)² – 2 (1) = 4

Slope of normal = – 1/Slope of tangent = – 1/4

Given, (x₁, y₁) = (1, 4), 

The equation of normal is,

y – y₁ = m (x – x₁)

y – 4 = -1/4 (x – 1)

4y – 16 = – x + 1

x + 4y = 17

问题 3. 求下列曲线在指定点的切线和法线方程：

(i) y = x ⁴ - bx ³ + 13x ² - 10x + 5 在 (0, 5)

解决方案：

We have,

y = x⁴ − bx³ + 13x² − 10x + 5

On differentiating both sides w.r.t. x, we get

dy/dx = 4x³ – 3bx² + 26x – 10

Slope of tangent, m= $\left( \frac{dy}{dx} \right)_{\left( 0, 5 \right)}$ = -10

Given, (x₁, y₁) = (0, 5) 

The equation of tangent is,

y – y₁ = m (x – x₁)

y – 5 = – 10 (x – 0)

y – 5 = -10x

y + 10x – 5 = 0

The equation of normal is,

y – y₁ = -1/m (x – x₁)

y – 5 = 1/10 (x – 0)

10y – 50 = x

x – 10y + 50 = 0

(ii) y = x ⁴ - 6x ³ + 13x ² - 10x + 5 在 x = 1

解决方案：

We have,

y = x⁴ − 6x³ + 13x² − 10x + 5

When x = 1, we have y = 1 – 6 + 13 – 10 + 5 = 3

So, (x₁, y₁) = (1, 3) 

Now, y = x⁴ − 6x³ + 13x² − 10x + 5

On differentiating both sides w.r.t. x, we get

dy/dx = 4 x³ – 18 x² + 26x – 10

Slope of tangent, m = $\left( \frac{dy}{dx} \right)_{\left( 1, 3 \right)}$ = 4 – 18 + 26 – 10 = 2

The equation of tangent is,

y – y₁ = 2 (x – x₁)

y – 3 = 2 (x – 1)

y – 3 = 2x – 2

2x – y + 1 = 0

The equation of normal is,

y – y₁ = -1/m (x – x₁)

y – 3 = -1/2 (x – 1)

2y – 6 = – x + 1

x + 2y – 7 = 0

(iii) y = x ²在 (0, 0)

解决方案：

We have,

y = x²

On differentiating both sides w.r.t. x, we get

dy/dx = 2x

Given, (x₁, y₁) = (0, 0) 

Slope of tangent, m= $\left(\frac{dy}{dx} \right)_{\left( 0, 0 \right)}$ = 2 (0) = 0

The equation of tangent is, 

y – y₁ = m (x – x₁)

y – 0 = 0 (x – 0)

y = 0

The equation of normal is,

y – y₁ = -1/m (x – x₁)

y – 0 = -1/0 (x – 0)

x = 0

(iv) y = 2x ² - 3x - 1 在 (1, -2)

解决方案：

We have,

y = 2x² − 3x − 1

On differentiating both sides w.r.t. x, we get

dy/dx = 4x – 3

Given, (x₁, y₁) = (1, -2) 

Slope of tangent, m = $\left( \frac{dy}{dx} \right)_{\left( 1, - 2 \right)}$ = 4 – 3 = 1

The equation of tangent is,

y – y₁ = m (x – x₁)

y + 2 = 1 (x – 1)

y + 2 = x – 1

x – y – 3 = 0

The equation of normal is,

y – y₁ = -1/m (x – x₁)

y + 2 = -1 (x – 1)

y + 2 = -x + 1

x + y + 1 = 0

(五) $y^2 = \frac{x^3}{4 - x}$在 (2, -2)

解决方案：

We have,

$y^2 = \frac{x^3}{4 - x}$

On differentiating both sides w.r.t. x, we get

$2y \frac{dy}{dx} = \frac{\left( 4 - x \right)\left( 3 x^2 \right) - x^3 \left( - 1 \right)}{\left( 4 - x \right)^2}$

= $\frac{12 x^2 - 3 x^3 + x^3}{\left( 4 - x \right)^2}$

= $\frac{12 x^2 - 2 x^3}{\left( 4 - x \right)^2}$

Given, (x₁, y₁) = (2, -2)

Slope of tangent, m= $\left( \frac{dy}{dx} \right)_{\left( 2, - 2 \right)}$ = $\frac{48 - 16}{- 16}$ = -2

The equation of tangent is,

y – y₁ = m (x – x₁)

y + 2 = -2 (x – 2)

y + 2 = -2x + 4

2x + y – 2 = 0

The equation of normal is,

y – y₁ = -1/m (x – x₁)

y + 2 = 1/2 (x – 2)

2y + 4 = x – 2

x – 2y – 6 = 0

(vi) y = x ² + 4x + 1 在 x = 3

解决方案：

We have,

y = x² + 4x + 1

On differentiating both sides w.r.t. x, we get,

dy/dx = 2x + 4

When x = 3, y = 9 + 12 + 1 = 22

So, (x₁, y₁) = (3, 22) 

Slope of tangent, m= $\left( \frac{dy}{dx} \right)_{x = 3}$ = 10

The equation of tangent is,

y – y₁ = m (x – x₁)

y – 22 = 10 (x – 3)

y – 22 = 10x – 30

10x – y – 8 = 0

The equation of normal is,

y – y₁ = -1/m (x – x₁)

y – 22 = -1/10 (x – 3)

10y – 220 = – x + 3

x + 10y – 223 = 0

(七) $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$在 (a cos θ, b sin θ)

解决方案：

We have,

$\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$

On differentiating both sides w.r.t. x, we get

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

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

dy/dx = -x b²/y a²

Slope of tangent, m= $\left( \frac{dy}{dx} \right)_{\left( a \cos \theta, b \sin \theta \right)} =\frac{- a \cos \theta \left( b^2 \right)}{b \sin \theta \left( a^2 \right)}=\frac{- b \cos \theta}{a \sin \theta}$

Given, (x₁, y₁) = (a cos θ, b sin θ)

The equation of tangent is,

y – y₁ = m (x – x₁)

y – b sin θ = -bcosθ/asinθ  (x – a cos θ)

ay sin θ – ab sin² θ = -bx cos θ + ab cos² θ

bx cos θ + ay sin θ = ab

On dividing by ab, we get

x/a cosθ + y/b sinθ = 1

The equation of normal is,

y – y₁ = -1/m (x – x₁)

y – b sin θ = asinθ/bcosθ  (x – a cos θ)

by cos θ – b² sin θ cos θ = ax sin θ – a² sin θ cos θ

ax sin θ – by cos θ = (a² – b²) sin θ cos θ

On dividing by sin θ cos θ, we get

ax sec θ – by cosec θ = a² – b²

(八) $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$在 (a sec θ, b tan θ)

解决方案：

We have,

$\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$

On differentiating both sides w.r.t. x, we get

$\frac{2x}{a^2} - \frac{2y}{b^2}\frac{dy}{dx} = 0$

$\frac{2y}{b^2}\frac{dy}{dx} = \frac{2x}{a^2}$

dy/dx = x b²/y a²

Slope of tangent}, m= $\left(\frac{dy}{dx} \right)_{\left( a \sec \theta, b \tan \theta \right)} =\frac{a \sec \theta \left( b^2 \right)}{b \tan \theta \left( a^2 \right)}=\frac{b}{a \sin \theta}$

Given, (x₁, y₁) = (a sec θ, b tan θ)

The equation of tangent is,

y – y₁ = m (x – x₁)

y – b tan θ = $\frac{b}{a \sin \theta}$ (x – a sec θ)

ay sin θ – ab(sin² θ/cos θ) = bx – (ab/cos θ) 

$\frac{ay \sin \theta \cos \theta - ab \sin^2 \theta}{\cos \theta} = \frac{bx \cos \theta - ab}{\cos \theta}$

ay sin θ cos θ – ab sin² θ = bx cos θ – ab

bx cos θ – ay sin θ cos θ = ab (1 – sin² θ)

bx cos θ – ay sin θ cos θ = ab cos² θ

On dividing by ab cos² θ, we get

x/a sec θ – y/b tan θ = 1

The equation of normal is,

y – y₁ = -1/m (x – x₁)

y – b tan θ = -a sin θ/b (x – a sec θ)

by – b² tan θ = -ax sin θ + a² tan θ

ax sin θ + by = (a² + b²) tan θ

On dividing by tan θ, we get

ax cos θ + by cot θ = a² + b²

(ix) y ² = 4ax at (a/m ² , 2a/m)

解决方案：

We have,

y² = 4ax

On differentiating both sides w.r.t. x, we get

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

dy/dx = 2a/y

Given, (x₁, y₁) = (a/m², 2a/m)

Slope of tangent = $\left( \frac{dy}{dx} \right)_{\left( \frac{a}{m^2}, \frac{2a}{m} \right)} =\frac{2a}{\left( \frac{2a}{m} \right)}$ = m

The equation of tangent is, 

y – y₁ = m (x – x₁)

$y - \frac{2a}{m} = m \left( x - \frac{a}{m^2} \right)$

$\frac{my - 2a}{m} = m\left( \frac{m^2 x - a}{m^2} \right)$

my – 2a = m² x – a

m² x – my + a = 0

The equation of normal is,

y – y₁ = -1/m (x – x₁)

$y - \frac{2a}{m} = \frac{- 1}{m}\left( x - \frac{a}{m^2} \right)$

$\frac{my - 2a}{m} = \frac{- 1}{m}\left( \frac{m^2 x - a}{m^2} \right)$

m³ y – 2a m² = – m² x + a

m² x + m³ y – 2a m² – a = 0

(x) c ² (x ² + y ² ) = x ² y ²在 (c/cos θ, c/sin θ)

解决方案：

We have,

c² (x² + y²) = x²y²

On differentiating both sides w.r.t. x, we get

2x c² + 2y c²(dy/dx) = x² 2y(dy/dx) + 2x y²

dy/dx(2y c² – 2 x² y) = 2x y² – 2x c²

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

Slope of tangent, m= $\left( \frac{dy}{dx} \right)_{\left( \frac{c}{\cos \theta}, \frac{c}{\sin \theta} \right)}$

= $\frac{\frac{c^3}{\cos \theta \sin^2 \theta} - \frac{c^3}{\cos \theta}}{\frac{c^3}{\sin\theta} - \frac{c^3}{\cos^2 \theta \sin\theta}}$

= $\frac{\frac{1 - \sin^2 \theta}{\cos\theta \sin^2 \theta}}{\frac{\cos^2 \theta - 1}{\cos^2 \theta \sin\theta}}$

= $\frac{co s^2 \theta}{\cos \theta \sin^2 \theta} \times \frac{\cos^2 \theta \sin\theta}{- \sin^2 \theta}$

= -cos³ θ/ sin³ θ 

Given, (x₁, y₁) = (c/cos θ, c/sin θ)

The equation of tangent is,

y – y₁ = m (x – x₁)

$y - \frac{c}{\sin \theta} = \frac{- \cos^3 \theta}{\sin^3 \theta} \left( x - \frac{c}{\cos \theta} \right)$

$\frac{y\sin\theta - c}{\sin\theta} = \frac{- \cos^3 \theta}{\sin^3 \theta}\left( \frac{x \cos\theta - c}{\cos\theta} \right)$

sin² θ (y sin θ – c) = -cos² θ (x cos θ – c)

y sin³ θ – c sin² θ = – x cos³ θ + c cos² θ

x cos³ θ + y sin³ θ = c ( sin² θ + cos² θ)

x cos³ θ + y sin³ θ = c

The equation of normal is,

y – y₁ = -1/m (x – x₁)

$y - \frac{c}{\sin \theta} = \frac{\sin^3 \theta}{\cos^3 \theta}\left( x - \frac{c}{\cos \theta} \right)$

$\cos^3 \theta\left( y - \frac{c}{\sin \theta} \right) = \sin^3 \theta\left( x - \frac{c}{\cos \theta} \right)$

$y \cos^3 \theta - \frac{c \cos^3 \theta}{\sin\theta} = x \sin^3 \theta - \frac{c \sin^3 \theta}{\cos\theta}$

$x \sin^3 \theta - y \cos^3 \theta = \frac{c \sin^3 \theta}{\cos\theta} - \frac{c \cos^3 \theta}{\sin\theta}$

$x \sin^3 \theta - y \cos^3 \theta = c\left( \frac{\sin^4 \theta - \cos^4 \theta}{\cos\theta \sin\theta} \right)$

$x \sin^3 \theta - y \cos^3 \theta = c\left[ \frac{\left( \sin^2 \theta + \cos^2 \theta \right)\left( \sin^2 \theta - \cos^2 \theta \right)}{\cos\theta \sin\theta} \right]$

$\sin^3 \theta - y \cos^3 \theta =2c \left[ \frac{- \left( \cos^2 \theta - \sin^2 \theta \right)}{2\cos\theta \sin\theta} \right]$

sin³ θ – ycos³ θ = 2c[-cos (2θ)/sin(2θ)]

sin³ θ – y cos³ θ = -2c cot 2θ

sin³ θ – y cos³ θ + 2c cot 2θ = 0

(xi) xy = c ²在 (ct, c/t)

解决方案：

We have,

xy = c²

On differentiating both sides w.r.t. x, we get

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

dy/dx = – y/x

Given, (x₁, y₁) = (ct, c/t)

Slope of tangent, m= $\left( \frac{dy}{dx} \right)_{\left( ct, \frac{c}{t} \right)} =\frac{- \frac{c}{t}}{ct}=\frac{- 1}{t^2}$

The equation of tangent is,

y – y₁ = m (x – x₁)

$y - \frac{c}{t} = \frac{- 1}{t^2} \left( x - ct \right)$

$\frac{yt - c}{t} = \frac{- 1}{t^2} \left( x - ct \right)$

yt² – ct = -x + ct

x + y t² = 2ct

The equation of normal is,

y – y₁ = -1/m (x – x₁)

y – c/t – t²(x – ct)

yt – c = t³ x – c t⁴

x t³ – yt = c t⁴ – c

(十二) $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$在 (x ₁ , y ₁ )

解决方案：

We have,

$\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$

On differentiating both sides w.r.t. x,

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

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

dy/dx = – x b²/y a²

Slope of tangent, m= $\left( \frac{dy}{dx} \right)_{\left( x_1 , y_1 \right)} =\frac{- x_1 b^2}{y_1 a^2}$

The equation of tangent is,

y – y₁ = m (x – x₁)

y – y₁ = – x₁ b²/y₁ a²(x – x₁)

y y₁ a² – y₁² a² = -x x₁ b² + x₁² b²

x x₁ b² + y y₁ a² = x₁² b² + y₁² a₂  . . . . (1)

Given (x₁, y₁) lies on the curve, we get

$\frac{{x_1}^2}{a^2} + \frac{{y_1}^2}{b^2} = 1$

$\frac{{x_1}^2 b^2 + {y_1}^2 a^2}{a^2 b^2} = 1$

x₁² b² + y₁² a² = a² b²

Substituting this in (1), we get

x x₁ b² + y y₁ a² = a² b²

On dividing this by a² b², we get

$\frac{x x_1}{a^2} + \frac{y y_1}{b^2} = 1$

The equation of normal is,

y – y₁ = m (x – x₁)

y – y₁ = y₁ a²/x₁ b²(x – x₁)

y x₁ b² – x₁ y₁ b² = x y₁ a² – x₁ y₁ a²

x y₁ a² – y x₁ b² = x₁ y₁ a² – x₁ y₁ b²

x y₁ a² – y x₁ b² = x₁ y₁ (a² – b²)

On dividing by x₁ y₁, we get

$\frac{a^2 x}{x_1} - \frac{b^2 y}{y_1} = a^2 - b^2$

(十三) $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$在 (x ₀ , y ₀ )

解决方案：

We have,

$\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$

On differentiating both sides w.r.t. x, we get

$\frac{2x}{a^2} - \frac{2y}{b^2}\frac{dy}{dx} = 0$

$\frac{2y}{b^2}\frac{dy}{dx} = \frac{2x}{a^2}$

dy/dx = x b²/y a²

Slope of tangent, m= $\left( \frac{dy}{dx} \right)_{\left( x_0 , y_0 \right)} =\frac{x_0 b^2}{y_0 a^2}$

The equation of tangent is,

y – y₁ = m (x – x₁)

y – y₀ = x₀ b²/y₀ a²(x – x₀)

y y₀ a² – y₀² a² = x x₀ b² – x₀² b²

x x₀ b² – y y₀ a² = x₀² b² – y₀² a²  . . . . (1)

$\frac{{x_0}^2}{a^2} - \frac{{y_0}^2}{b^2} = 1$

x₀² b² – y₀² a² = a² b² 

Substituting this in eq(1), we get,

x x₀ b² – y y₀ a² = a² b²

Dividing this by a² b², we get

$\frac{x x_0}{a^2} - \frac{y y_0}{b^2} = 1$

The equation of normal is,

y – y₁ = m (x – x₁)

y – y₀ = y₀ a²/x₀ b²(x – x₀)

y x₀ b² – x₀ y₀ b² = -x y₀ a² + x₀ y₀ a² 

x y₀ a² + y x₀ b² = x₀ y₀ a² + x₀ y₀ b²

x y₀ a² + y x₀ b² = x₀ y₀ (a² + b²)

Dividing by x₀ y₀, we get

$\frac{a^2 x}{x_0} + \frac{b^2 y}{y_0} = a^2 + b^2$

(十四) $x^\frac{2}{3} + y^\frac{2}{3} = 2$在 (1, 1)

解决方案：

We have,

$x^\frac{2}{3} + y^\frac{2}{3} = 2$

On differentiating both sides w.r.t. x, we get

$\frac{2}{3} x^\frac{- 1}{3} + \frac{2}{3} y^\frac{- 1}{3} \frac{dy}{dx} = 0$

$\frac{dy}{dx} = \frac{- x^\frac{- 1}{3}}{y^\frac{- 1}{3}} = \frac{- y^\frac{1}{3}}{x^\frac{1}{3}}$

Slope of tangent, m= $\left( \frac{dy}{dx} \right)_{\left( 1, 1 \right)}$= -1

Given, (x₁, y₁) = (1, 1)

The equation of tangent is,

y – y₁ = m (x – x₁)

y – 1 = -1 (x – 1)

y – 1 = -x + 1

x + y – 2 = 0

The equation of normal is,

y – y₁ = -1/m (x – x₁)

y – 1 = 1 (x – 1)

y – 1 = x – 1

y – x = 0

(xv) x ² = 4y 在 (2, 1)

解决方案：

We have,

x² = 4y 

On differentiating both sides w.r.t. x, we get

2x = 4dy/dx

dy/dx = x/2

Slope of tangent, m= $\left( \frac{dy}{dx} \right)_{\left( 2, 1 \right)}$ = 2/2 = 1

Given, (x₁, y₁) = (2, 1)

The equation of tangent is,

y – y₁ = m (x – x₁)

y – 1 = 1 (x – 2)

y – 1 = x – 2

x – y – 1 = 0

The equation of normal is,

y – y₁ = -1/m (x – x₁)

y – 1 = – 1 (x – 2)

y – 1 = – x + 2

x + y – 3 = 0

(xvi) y ² = 4x 在 (1, 2)

解决方案：

We have,

y² = 4x

On differentiating both sides w.r.t. x, we get

2y (dy/dx) = 4

dy/dx = 2/y

Slope of tangent, m= $\left( \frac{dy}{dx} \right)_{\left( 1, 2 \right)}$= 2/2 = 1

Given, (x₁, y₁) = (1, 2)

The equation of tangent is,

y – y₁ = m (x – x₁)

y – 2 = 1 (x – 1)

y – 2 = x – 1

x – y + 1 = 0

The equation of normal is,

y – y₁ = -1/m (x – x₁)

y – 2 = -1 (x – 1)

y – 2 = -x + 1

x + y – 3 = 0

(xvii) 4x ² + 9y ² = 36 at (3 cos θ, 2 sin θ)

解决方案：

We have,

4x² + 9y² = 36

On differentiating both sides w.r.t. x, we get

8x + 18y dy/dx = 0

18y dy/dx = – 8x

dy/dx = -8x/18y = -4x/9y

Slope of tangent, m = $\left( \frac{dy}{dx} \right)_{\left( 3 \cos\theta, 2 \sin\theta \right)} =\frac{- 12\cos\theta}{18\sin\theta}=\frac{- 2 \cos\theta}{3 \sin\theta}$

The equation of tangent is,

y – y₁ = m (x – x₁)

y – 2 sin θ = -2 cos θ/3 sin θ(x – 3 cos θ)

3y sin θ – 6 sin² θ = -2x cos θ + 6 cos² θ

2x cos θ + 3y sin θ = 6 (cos² θ + sin² θ)

2x cos θ + 3y sin θ = 6

The equation of normal is,

y – y₁ = -1/m (x – x₁)

y – 2 sin θ = -3 sin θ/2 cos θ(x – 3 cos θ)

2y cos θ – 4 sin θ cos θ = 3x sin θ – 9 sin θ cos θ

3x sin θ – 2y cos θ – 5sin θ cos θ = 0

(xviii) y ² = 4ax at (x ₁ , y ₁ )

解决方案：

We have,

y² = 4ax

On differentiating both sides w.r.t. x, we get

2y dy/dx = 4a

dy/dx = 2a/y

At (x₁, y₁), we have

Slope of tangent = $\left( \frac{dy}{dx} \right)_{\left( x_1 , y_1 \right)} =\frac{2a}{y_1}$= m

The equation of tangent is,

y – y₁ = m (x – x₁)

$y - y_1 = \frac{2a\left( x - x_1 \right)}{y_1}$

y y₁ – y₁² = 2ax – 2a x₁

y y₁ – 4a x₁ = 2ax – 2a x₁

y y₁ = 2ax + 2a x₁

y y₁ = 2a (x + x₁)

The equation of normal is,

y – y₁ = -1/m (x – x₁)

y – y₁ = -y₁/2a (x – x₁)

(十九) $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$在 (√2a, b)

解决方案：

We have,

$\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$

On differentiating both sides w.r.t. x, we get

$\frac{2x}{a^2} - \frac{2y}{b^2}\frac{dy}{dx} = 0$

$\frac{2y}{b^2}\frac{dy}{dx} = \frac{2x}{a^2}$

dy/dx = x b²/y a²

Slope of tangent, m= $\left( \frac{dy}{dx} \right)_{\left( \sqrt{2}a,b \right)} =\frac{\sqrt{2}a b^2}{b a^2}=\frac{\sqrt{2}b}{a}$

The equation of tangent is,

y – y₁ = m (x – x₁)

y – b = √2b/a(x – √2a)

ay – ab = √2 bx – 2ab

√2 bx – ay = ab

$\frac{\sqrt{2}x}{a} - \frac{y}{b} = 1$

The equation of normal is, 

y – y₁ = -1/m (x – x₁)

y – b = – a/√2b(x – √2a)

√2 by – √2 b² = – ax + √2 a² 

ax + √2 by = √2 b² + √2 a²

ax/√2 + by = a² + b²

问题 4. 求曲线 x = θ + sin θ, y = 1 + cos θ 在 θ = π/4 处的切线方程。

解决方案：

We have,

x = θ + sin θ, y = 1 + cos θ

$\frac{dx}{d\theta} = 1 + \cos \theta$ and $\frac{dy}{d\theta} = - \sin \theta$

$\frac{dy}{dx} = \frac{\frac{dy}{d\theta}}{\frac{dx}{d\theta}} = \frac{- \sin \theta}{1 + \cos \theta}$

Slope of tangent, m = $\left( \frac{dy}{dx} \right)_{\theta = \frac{\pi}{4}}$

= $\frac{- \sin \frac{\pi}{4}}{1 + \cos \frac{\pi}{4}}$

= $\frac{\frac{- 1}{\sqrt{2}}}{1 + \frac{1}{\sqrt{2}}}$

= $\frac{-1}{\sqrt{2} + 1}$

= $\frac{-1}{\sqrt{2} + 1}\times\frac{\sqrt{2} - 1}{\sqrt{2} - 1}$

= 1 – √2 

Given, (x₁, y₁) = (π/4 + sin π/4, 1 + cos π/4) = (π/4 + 1/√2, 1 + 1/√2)

The equation of tangent is,

y – y₁ = m (x – x₁)

y – (1 + 1/√2) = (1 – √2) [x – (π/4 + 1/√2)]

y – 1 – 1/√2 = (1 – √2) (x – π/4 – 1/√2)

问题 5. 求下列曲线在指定点处的切线和法线方程。

(i) x = θ + sin θ, y = 1 + cos θ 在 θ = π/2

解决方案：

We have,

x = θ + sin θ and y = 1 + cos θ

dx/dθ = 1 + cos θ and dy/dθ = -sinθ

$\frac{dy}{dx} = \frac{\frac{dy}{d\theta}}{\frac{dx}{d\theta}}$

= $\frac{- \sin\theta}{1 + \cos\theta}$

Slope of tangent, m= $\left( \frac{dy}{dx} \right)_{\theta = \frac{\pi}{2}} =\frac{- \sin\frac{\pi}{2}}{1 + \cos\frac{\pi}{2}}$

= -1/(1 + 0)

= -1

Given, (x₁, y₁) = (π/2 + sin π/2, 1 + cos π/2) = (π/2 + 1, 1)

The equation of tangent is,

y – y₁ = m (x – x₁)

y – 1 = -1 (x – π/2 – 1)

2y – 2 = – 2x + π + 2

x + 2y – π – 4 = 0

The equation of normal is,

y – y₁ = -1/m (x – x₁)

y – 1 = 1 (x – π/2 -1)

2y – 2 = 2x – π – 2

2x – 2y = π

(二) $x = \frac{2 a t^2}{1 + t^2}, y = \frac{2 a t^3}{1 + t^2}$在 t = 1/2

解决方案：

We have,

$x = \frac{2 a t^2}{1 + t^2}, y = \frac{2 a t^3}{1 + t^2}$

dx/dt = $\frac{\left( 1 + t^2 \right)\left( 4at \right) - 2a t^2 \left( 2t \right)}{\left( 1 + t^2 \right)^2}$

= $\frac{4at}{\left( 1 + t^2 \right)^2}$

dy/dt = $\frac{\left( 1 + t^2 \right)6a t^2 - 2a t^3 \left( 2t \right)}{\left( 1 + t^2 \right)^2}$

 = $\frac{6a t^2 + 2a t^4}{\left( 1 + t^2 \right)^2}$

$\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{\frac{6a t^2 + 2a t^4}{\left( 1 + t^2 \right)^2}}{\frac{4at}{\left( 1 + t^2 \right)^2}} = \frac{6a t^2 + 2a t^4}{4at}$

Slope of tangent, m= $\left( \frac{dy}{dx} \right)_{t = \frac{1}{2}} =\frac{\frac{3a}{2} + \frac{a}{8}}{2a}=\frac{13a}{8}(\frac{1}{2a})=\frac{13}{16}$

Given, (x₁, y₁) = $\left(\frac{2 a t^2}{1 + t^2},\frac{2 a t^3}{1 + t^2}\right)=\left( \frac{\frac{a}{2}}{1 + \frac{1}{4}}, \frac{\frac{a}{4}}{1 + \frac{1}{4}} \right) = \left( \frac{\frac{a}{2}}{\frac{5}{4}}, \frac{\frac{a}{4}}{\frac{5}{4}} \right) = \left( \frac{2a}{5}, \frac{a}{5} \right)$

The equation of tangent is,

y – y₁ = m (x – x₁)

$y - \frac{a}{5} = \frac{13}{16}\left( x - \frac{2a}{5} \right)$

$\frac{5y - a}{5} = \frac{13}{16}\left( \frac{5x - 2a}{5} \right)$

$5y - a = \frac{13}{16}\left( 5x - 2a \right)$

80y – 16a = 65x – 26a

65x – 80y – 10a = 0

13x – 16y – 2a = 0

The equation of normal is,

y – y₁ = -1/m (x – x₁)

$y - \frac{a}{5} = \frac{- 16}{13} \left( x - \frac{2a}{5} \right)$

$\frac{5y - a}{5} = \frac{- 16}{13}\left( \frac{5x - 2a}{5} \right)$

$5y - a = \frac{- 16}{13}\left( 5x - 2a \right)$

65y – 13a = – 80x + 32a

80x + 65y – 45a = 0

16x + 13y – 9a = 0

(iii) x = 在² ，y = 2at 在 t = 1

解决方案：

We have,

x = at², y = 2at

dx/dt = 2at and dy/dt = 2a

$\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{2a}{2at} = \frac{1}{t}$

Slope of tangent, m= $\left( \frac{dy}{dx} \right)_{t = 1}$ = 1

Given, (x₁ , y₁) = (a, 2a)

The equation of tangent is,

y – y₁ = m (x – x₁)

y – 2a = 1 (x – a)

y – 2a = x – a

x – y + a = 0

Equation of normal:

y – y₁ = -1/m (x – x₁)

y – 2a = – 1 (x – a)

y – 2a = – x + a

x + y = 3a

(iv) x = a sec t, y = b tan t 在 t

解决方案：

We have,

x = a sec t, y = b tan t

dx/dt = a sect tant and dy/dt = b sec²t

$\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{b \sec^2 t}{a \sec t \tan t} = \frac{b}{a}\cosec t$

Slope of tangent, m = $\left( \frac{dy}{dx} \right)_{t = t} =\frac{b}{a}\cosec t$

Given (x₁, y₁) = (a sec t, b tan t)

The equation of tangent is,

y – y₁ = m (x – x₁)

y – b tan t = (b/a) cosec t (x – a sec t)

$y - \frac{b \sin t}{\cos t} = \frac{b}{a \sin t}\left( x - \frac{a}{\cos t} \right)$

$\frac{y \cos t - b \sin t}{\cos t} = \frac{b}{a \sin t}\left( \frac{x \cos t - a}{\cos t} \right)$

$y \cos t - b \sin t = \frac{b}{a \sin t}\left( x \cos t - a \right)$

ay sin t cos t – ab sin² t = bx cos t – ab

bx cos t – ay sin t cos t – ab (1 – sin² t) = 0

bx cos t – ay sin t cos t = ab cos² t

On dividing by cos² t, we get

bx sec t – ay tan t = ab

The equation of normal is,

y – y₁ = -1/m (x – x₁)

y – b tant = -a/b sint(x – asect)

$y - b \frac{\sin t}{\cos t} = \frac{- a}{b}\sin t\left( x - \frac{a}{\cos t} \right)$

$\frac{y \cos t - b \sin t}{\cos t} = \frac{- a}{b}\sin t\left( \frac{x \cos t - a}{\cos t} \right)$

ycost − bsint = − a/b​sint(xcost − a)

by cos t – b² sin t = – ax sin t cos t + a² sin t

ax sin t cos t + by cos t = (a² + b²) sin t

On dividing both sides by sin t, we get

ax cos t + by cot t = a² + b²

(v) x = a(θ + sin θ), y = a(1 - cos θ) 在 θ

解决方案：

We have,

x = a(θ + sin θ), y = a(1 − cos θ)

dx/dθ = a(1 + cosθ) and dy/dθ = asinθ

$\frac{dy}{dx} = \frac{\frac{dy}{d\theta}}{\frac{dx}{d\theta}}$

= $\frac{a\sin\theta}{a\left( 1 + \cos\theta \right)}$

= $\frac{\sin\theta}{\left( 1 + \cos\theta \right)}$

= $\frac{2\sin\frac{\theta}{2}\cos\frac{\theta}{2}}{2 \cos^2 \frac{\theta}{2}}$ 

= tan θ/2   . . . . (1)

Slope of tangent, m= $\left( \frac{dy}{dx} \right)_\theta =\tan\frac{\theta}{2}$

Given (x₁, y₁) = [a(θ + sin θ), a(1 − cos θ)]

The equation of tangent is,

y – y₁ = m (x – x₁)

y – a (1 – cos θ) = tan θ/2 [x – a (θ + sin θ)]

y − a(2sin²θ/2​) = xtanθ/2​ − aθtanθ/2 ​− atanθ/2​sinθ

$y - a\left( 2 \sin^2 \frac{\theta}{2} \right) = x\tan\frac{\theta}{2} - a\theta\tan\frac{\theta}{2} - a\frac{2\sin\frac{\theta}{2}\cos\frac{\theta}{2}}{2 \cos^2 \frac{\theta}{2}}2\sin\frac{\theta}{2}\cos\frac{\theta}{2}$

y − 2asin²θ/2 ​ =(x − aθ)tan θ/2 − 2asin²θ/2​

y = (x – aθ) tan θ/2

The equation of normal is,

y – a (1 – cos θ) = -cot θ/2 [x – a (θ + sin θ)]

tanθ/2​[y − a(2sin²θ/2​)] = −x + aθ + asinθ

tanθ/2​[y − a{2(1 − cos²θ/2​)}] = −x + aθ + asinθ

tan θ/2 (y – 2a) + a (2sin θ/2 cosθ/2 = -x + aθ + a sinθ

tan θ/2 (y – 2a) + a sin θ = -x + aθ + a sin θ

tan θ/2 (y – 2a) = – x + aθ

tan θ/2 (y – 2a) + x – θ = 0

(vi) x = 3 cos θ - cos ³ θ, y = 3 sin θ - sin ³ θ

解决方案：

We have,

x = 3 cos θ − cos³ θ, y = 3 sin θ − sin³ θ

dx/dθ = -3sin θ + 3 cos²θ sin θ and dy/dθ = 3 cos θ – 3 sin²θ cos θ

$\frac{dy}{dx}=\frac{\frac{dy}{d \theta}}{\frac{dx}{d \theta}}$

= $\frac{3\cos \theta-3\sin^2 \theta\cos \theta}{-3\sin \theta+3\cos^2\theta \sin \theta}$

= $\frac{\cos \theta(1-\sin^2 \theta}{-\sin \theta(1-\cos^2 \theta)}$

=  cos³ θ/ -sin³ θ 

= tan³ θ

So the equation of the tangent at θ is,

y – 3 sin θ + sin³ θ = -tan³ θ (x – 3 cos θ + cos³ θ)

4 (y cos³ θ – x sin³ θ) = 3 sin 4θ

So the equation of normal at θ is,

y – 3 sin θ + sin³ θ= (1/tan³ θ) (x – 3 cos θ + cos³ θ)

sin³ θ – x cos³ θ = 3sin⁴ θ – sin⁶ θ – 3cos⁴ θ + cos⁶ θ

问题 6. 求曲线 x ² + 2y ² - 4x - 6y + 8 = 0 在横坐标为 2 的点的法线方程？

解决方案：

Given that abscissa = 2. i.e., x = 2

x² + 2y² − 4x − 6y + 8 = 0  . . . . (1)

On differentiating both sides w.r.t. x, we get

2x + 4y dy/dx – 4 – 6 dy/dx = 0

dy/dx(4y – 6) = 4 – 2x

$\frac{dy}{dx} = \frac{4 - 2x}{4y - 6} = \frac{2 - x}{2y - 3}$

When x = 2, we get

4 + 2y² – 8 – 6y + 8 = 0

2y² – 6y + 4 = 0

y² – 3y + 2 = 0

y = 2 or y = 1

m (tangent) at x = 2 is 0

Normal is perpendicular to tangent so, m₁m₂ = –1

m (normal) at x = 2 is 1/0, which is undefined.

The equation of normal is given by y – y₁ = m (normal) (x – x₁)

x = 2