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Textbook | NCERT |

Class | Class 12th |

Subject | Maths |

Chapter | 16 |

Exercise | 16.2 |

Category | RD Sharma Solutions |

**RD Sharma Class 12 Ex 16.2 Solutions Chapter 16 Tangents and **

### Question 1. Find the equation of the tangent to the curve √x + √y = a at the point (a^{2}/4, a^{2}/4).

**Solution:**

We have,

√x + √y = a

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

dy/dx = -√y/√x

Given, (x_{1}, y_{1}) = (a^{2}/4, a^{2}/4),

Slope of tangent, m =

The equation of tangent is,

y – y_{1} = m (x – x_{1})

y – a^{2}/4 = –1(x – a^{2}/4)

y – a^{2}/4 = –x + a^{2}/4

x + y = a^{2}/2

### Question 2. Find the equation of the normal to y = 2x^{3} − x^{2} + 3 at (1, 4).

**Solution:**

We have,

y = 2x

^{3}− x^{2}+ 3On differentiating both sides w.r.t. x, we get

dy/dx = 6x

^{2 }– 2xSlope of tangent = = 6 (1)

^{2}– 2 (1) = 4Slope of normal = – 1/Slope of tangent = – 1/4

Given, (x

_{1}, y_{1}) = (1, 4),The equation of normal is,

y – y

_{1}= m (x – x_{1})y – 4 = -1/4 (x – 1)

4y – 16 = – x + 1

x + 4y = 17

### Question 3. Find the equation of the tangent and the normal to the following curve at the indicated point:

### (i) y = x^{4} − bx^{3} + 13x^{2} − 10x + 5 at (0, 5)

**Solution:**

We have,

y = x

^{4}− bx^{3}+ 13x^{2}− 10x + 5On differentiating both sides w.r.t. x, we get

dy/dx = 4x

^{3}– 3bx^{2}+ 26x – 10Slope of tangent, m= = -10

Given, (x

_{1}, y_{1}) = (0, 5)The equation of tangent is,

y – y

_{1}= m (x – x_{1})y – 5 = – 10 (x – 0)

y – 5 = -10x

y + 10x – 5 = 0

The equation of normal is,

y – y

_{1}= -1/m (x – x_{1})y – 5 = 1/10 (x – 0)

10y – 50 = x

x – 10y + 50 = 0

### (ii) y = x^{4} − 6x^{3} + 13x^{2} − 10x + 5 at x = 1

**Solution:**

We have,

y = x

^{4}− 6x^{3}+ 13x^{2}− 10x + 5When x = 1, we have y = 1 – 6 + 13 – 10 + 5 = 3

So, (x

_{1}, y_{1}) = (1, 3)Now, y = x

^{4}− 6x^{3}+ 13x^{2}− 10x + 5On differentiating both sides w.r.t. x, we get

dy/dx = 4 x

^{3}– 18 x^{2}+ 26x – 10Slope of tangent, m = = 4 – 18 + 26 – 10 = 2

The equation of tangent is,

y – y

_{1}= 2 (x – x_{1})y – 3 = 2 (x – 1)

y – 3 = 2x – 2

2x – y + 1 = 0

The equation of normal is,

y – y

_{1}= -1/m (x – x_{1})y – 3 = -1/2 (x – 1)

2y – 6 = – x + 1

x + 2y – 7 = 0

### (iii) y = x^{2} at (0, 0)

**Solution:**

We have,

y = x

^{2}On differentiating both sides w.r.t. x, we get

dy/dx = 2x

Given, (x

_{1}, y_{1}) = (0, 0)Slope of tangent, m= = 2 (0) = 0

The equation of tangent is,

y – y

_{1}= m (x – x_{1})y – 0 = 0 (x – 0)

y = 0

The equation of normal is,

y – y

_{1}= -1/m (x – x_{1})y – 0 = -1/0 (x – 0)

x = 0

### (iv) y = 2x^{2} − 3x − 1 at (1, −2)

**Solution:**

We have,

y = 2x

^{2 }− 3x − 1On differentiating both sides w.r.t. x, we get

dy/dx = 4x – 3

Given, (x

_{1}, y_{1}) = (1, -2)Slope of tangent, m = = 4 – 3 = 1

The equation of tangent is,

y – y

_{1}= m (x – x_{1})y + 2 = 1 (x – 1)

y + 2 = x – 1

x – y – 3 = 0

The equation of normal is,

y – y

_{1}= -1/m (x – x_{1})y + 2 = -1 (x – 1)

y + 2 = -x + 1

x + y + 1 = 0

### (v) at (2, -2)

**Solution:**

We have,

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

=

=

Given, (x_{1}, y_{1}) = (2, -2)

Slope of tangent, m= = = -2

The equation of tangent is,

y – y_{1} = m (x – x_{1})

y + 2 = -2 (x – 2)

y + 2 = -2x + 4

2x + y – 2 = 0

The equation of normal is,

y – y_{1} = -1/m (x – x_{1})

y + 2 = 1/2 (x – 2)

2y + 4 = x – 2

x – 2y – 6 = 0

### (vi) y = x^{2} + 4x + 1 at x = 3

**Solution:**

We have,

y = x

^{2}+ 4x + 1On differentiating both sides w.r.t. x, we get,

dy/dx = 2x + 4

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

So, (x

_{1}, y_{1}) = (3, 22)Slope of tangent, m= = 10

The equation of tangent is,

y – y

_{1}= m (x – x_{1})y – 22 = 10 (x – 3)

y – 22 = 10x – 30

10x – y – 8 = 0

The equation of normal is,

y – y

_{1}= -1/m (x – x_{1})y – 22 = -1/10 (x – 3)

10y – 220 = – x + 3

x + 10y – 223 = 0

### (vii) at (a cos θ, b sin θ)

**Solution:**

We have,

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

dy/dx = -x b^{2}/y a^{2}

Slope of tangent, m=

Given, (x_{1}, y_{1}) = (a cos θ, b sin θ)

The equation of tangent is,

y – y_{1} = m (x – x_{1})

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

ay sin θ – ab sin^{2} θ = -bx cos θ + ab cos^{2} 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 – y_{1} = -1/m (x – x_{1})

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

by cos θ – b^{2} sin θ cos θ = ax sin θ – a^{2} sin θ cos θ

ax sin θ – by cos θ = (a^{2} – b^{2}) sin θ cos θ

On dividing by sin θ cos θ, we get

ax sec θ – by cosec θ = a^{2} – b^{2}

### (viii) at (a sec θ, b tan θ)

**Solution:**

We have,

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

dy/dx = x b^{2}/y a^{2}

Slope of tangent}, m=

Given, (x_{1}, y_{1}) = (a sec θ, b tan θ)

The equation of tangent is,

y – y_{1} = m (x – x_{1})

y – b tan θ = (x – a sec θ)

ay sin θ – ab(sin^{2} θ/cos θ) = bx – (ab/cos θ)

ay sin θ cos θ – ab sin^{2} θ = bx cos θ – ab

bx cos θ – ay sin θ cos θ = ab (1 – sin^{2} θ)

bx cos θ – ay sin θ cos θ = ab cos^{2} θ

On dividing by ab cos^{2} θ, we get

x/a sec θ – y/b tan θ = 1

The equation of normal is,

y – y_{1} = -1/m (x – x_{1})

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

by – b^{2} tan θ = -ax sin θ + a^{2} tan θ

ax sin θ + by = (a^{2} + b^{2}) tan θ

On dividing by tan θ, we get

ax cos θ + by cot θ = a^{2} + b^{2}

### (ix) y^{2} = 4ax at (a/m^{2}, 2a/m)

**Solution:**

We have,

y^{2} = 4ax

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

dy/dx = 2a/y

Given, (x_{1}, y_{1}) = (a/m^{2}, 2a/m)

Slope of tangent = = m

The equation of tangent is,

y – y_{1} = m (x – x_{1})

my – 2a = m^{2} x – a

m^{2} x – my + a = 0

The equation of normal is,

y – y_{1} = -1/m (x – x_{1})

m^{3} y – 2a m^{2} = – m^{2} x + a

m^{2} x + m^{3} y – 2a m^{2} – a = 0

### (x) c^{2} (x^{2} + y^{2}) = x^{2}y^{2} at (c/cos θ, c/sin θ)

**Solution:**

We have,

c^{2} (x^{2} + y^{2}) = x^{2}y^{2}

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

2x c^{2} + 2y c^{2}(dy/dx) = x^{2} 2y(dy/dx) + 2x y^{2}

dy/dx(2y c^{2} – 2 x^{2} y) = 2x y^{2} – 2x c^{2}

Slope of tangent, m=

=

=

=

= -cos^{3} θ/ sin^{3} θ

Given, (x_{1}, y_{1}) = (c/cos θ, c/sin θ)

The equation of tangent is,

y – y_{1} = m (x – x_{1})

sin^{2 }θ (y sin θ – c) = -cos^{2} θ (x cos θ – c)

y sin^{3} θ – c sin^{2} θ = – x cos^{3} θ + c cos^{2} θ

x cos^{3} θ + y sin^{3} θ = c ( sin^{2} θ + cos^{2} θ)

x cos^{3} θ + y sin^{3} θ = c

The equation of normal is,

y – y_{1} = -1/m (x – x_{1})

sin^{3} θ – ycos^{3} θ = 2c[-cos (2θ)/sin(2θ)]

sin^{3} θ – y cos^{3} θ = -2c cot 2θ

sin^{3} θ – y cos^{3} θ + 2c cot 2θ = 0

### (xi) xy = c^{2} at (ct, c/t)

**Solution:**

We have,

xy = c^{2}

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

dy/dx = – y/x

Given, (x_{1}, y_{1}) = (ct, c/t)

Slope of tangent, m=

The equation of tangent is,

y – y_{1} = m (x – x_{1})

yt^{2} – ct = -x + ct

x + y t^{2} = 2ct

The equation of normal is,

y – y_{1} = -1/m (x – x_{1})

y – c/t – t^{2}(x – ct)

yt – c = t^{3} x – c t^{4}

x t^{3} – yt = c t^{4} – c

### (xii) at (x_{1}, y_{1})

**Solution:**

We have,

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

dy/dx = – x b^{2}/y a^{2}

Slope of tangent, m=

The equation of tangent is,

y – y_{1} = m (x – x_{1})

y – y_{1} = – x_{1} b^{2}/y_{1} a^{2}(x – x_{1})

y y_{1} a^{2} – y_{1}^{2} a^{2} = -x x_{1} b^{2} + x_{1}^{2} b^{2}

x x_{1} b^{2} + y y_{1} a^{2} = x_{1}^{2} b^{2} + y_{1}^{2} a_{2} . . . . (1)

Given (x_{1}, y_{1}) lies on the curve, we get

x_{1}^{2} b^{2} + y_{1}^{2} a^{2} = a^{2} b^{2}

Substituting this in (1), we get

x x_{1} b^{2} + y y_{1} a^{2} = a^{2} b^{2}

On dividing this by a^{2 }b^{2}, we get

The equation of normal is,

y – y_{1} = m (x – x_{1})

y – y_{1} = y_{1} a^{2}/x_{1} b^{2}(x – x_{1})

y x_{1} b^{2} – x_{1} y_{1} b^{2} = x y_{1} a^{2} – x_{1 }y_{1} a^{2}

x y_{1} a^{2} – y x_{1} b^{2} = x_{1} y_{1} a^{2} – x_{1} y_{1} b^{2}

x y_{1} a^{2} – y x_{1} b^{2} = x_{1} y_{1} (a^{2} – b^{2})

On dividing by x_{1} y_{1}, we get

### (xiii) at (x_{0} , y_{0})

**Solution:**

We have,

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

dy/dx = x b^{2}/y a^{2}

Slope of tangent, m=

The equation of tangent is,

y – y_{1} = m (x – x_{1})

y – y_{0} = x_{0} b^{2}/y_{0} a^{2}(x – x_{0})

y y_{0} a^{2} – y_{0}^{2} a^{2} = x x_{0} b^{2} – x_{0}^{2} b^{2}

x x_{0} b^{2} – y y_{0} a^{2} = x_{0}^{2} b^{2} – y_{0}^{2 }a^{2} . . . . (1)

x_{0}^{2}_{ }b^{2}_{ }– y_{0}^{2}_{ }a^{2} = a^{2 }b^{2 }

Substituting this in eq(1), we get,

x x_{0} b^{2} – y y_{0} a^{2} = a^{2 }b^{2}

Dividing this by a^{2 }b^{2}, we get

The equation of normal is,

y – y_{1} = m (x – x_{1})

y – y_{0} = y_{0} a^{2}/x_{0} b^{2}(x – x_{0})

y x_{0} b^{2} – x_{0 }y_{0} b^{2} = -x y_{0} a^{2} + x_{0} y_{0} a^{2}

x y_{0} a^{2} + y x_{0} b^{2} = x_{0} y_{0} a^{2} + x_{0} y_{0} b^{2}

x y_{0} a^{2} + y x_{0} b^{2} = x_{0} y_{0} (a^{2} + b^{2})

Dividing by x_{0} y_{0}, we get

### (xiv) at (1, 1)

**Solution:**

We have,

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

Slope of tangent, m= = -1

Given, (x_{1}, y_{1}) = (1, 1)

The equation of tangent is,

y – y_{1} = m (x – x_{1})

y – 1 = -1 (x – 1)

y – 1 = -x + 1

x + y – 2 = 0

The equation of normal is,

y – y_{1} = -1/m (x – x_{1})

y – 1 = 1 (x – 1)

y – 1 = x – 1

y – x = 0

### (xv) x^{2} = 4y at (2, 1)

**Solution:**

We have,

x

^{2}= 4yOn differentiating both sides w.r.t. x, we get

2x = 4dy/dx

dy/dx = x/2

Slope of tangent, m= = 2/2 = 1

Given, (x

_{1}, y_{1}) = (2, 1)The equation of tangent is,

y – y

_{1}= m (x – x_{1})y – 1 = 1 (x – 2)

y – 1 = x – 2

x – y – 1 = 0

The equation of normal is,

y – y

_{1}= -1/m (x – x_{1})y – 1 = – 1 (x – 2)

y – 1 = – x + 2

x + y – 3 = 0

### (xvi) y^{2} = 4x at (1, 2)

**Solution:**

We have,

y

^{2}= 4xOn differentiating both sides w.r.t. x, we get

2y (dy/dx) = 4

dy/dx = 2/y

Slope of tangent, m= = 2/2 = 1

Given, (x

_{1}, y_{1}) = (1, 2)The equation of tangent is,

y – y

_{1}= m (x – x_{1})y – 2 = 1 (x – 1)

y – 2 = x – 1

x – y + 1 = 0

The equation of normal is,

y – y

_{1}= -1/m (x – x_{1})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)

**Solution:**

We have,

4x

^{2}+ 9y^{2}= 36On 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 =

The equation of tangent is,

y – y

_{1}= m (x – x_{1})y – 2 sin θ = -2 cos θ/3 sin θ(x – 3 cos θ)

3y sin θ – 6 sin

^{2}θ = -2x cos θ + 6 cos^{2}θ2x cos θ + 3y sin θ = 6 (cos

^{2}θ + sin^{2 }θ)2x cos θ + 3y sin θ = 6

The equation of normal is,

y – y

_{1}= -1/m (x – x_{1})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})

**Solution:**

We have,

y^{2} = 4ax

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

2y dy/dx = 4a

dy/dx = 2a/y

At (x_{1}, y_{1}), we have

Slope of tangent = = m

The equation of tangent is,

y – y_{1} = m (x – x_{1})

y y_{1} – y_{1}^{2} = 2ax – 2a x_{1}

y y_{1} – 4a x_{1} = 2ax – 2a x_{1}

y y_{1} = 2ax + 2a x_{1}

y y_{1} = 2a (x + x_{1})

The equation of normal is,

y – y_{1} = -1/m (x – x_{1})

y – y_{1} = -y_{1}/2a (x – x_{1})

### (xix) at (√2a, b)

**Solution:**

We have,

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

dy/dx = x b^{2}/y a^{2}

Slope of tangent, m=

The equation of tangent is,

y – y_{1} = m (x – x_{1})

y – b = √2b/a(x – √2a)

ay – ab = √2 bx – 2ab

√2 bx – ay = ab

The equation of normal is,

y – y_{1} = -1/m (x – x_{1})

y – b = – a/√2b(x – √2a)

√2 by – √2 b^{2} = – ax + √2 a^{2 }

ax + √2 by = √2 b^{2} + √2 a^{2}

ax/√2 + by = a^{2} + b^{2}

### Question 4. Find the equation of the tangent to the curve x = θ + sin θ, y = 1 + cos θ at θ = π/4.

**Solution:**

We have,

x = θ + sin θ, y = 1 + cos θ

and

Slope of tangent, m =

=

=

=

=

= 1 – √2

Given, (x_{1}, y_{1}) = (π/4 + sin π/4, 1 + cos π/4) = (π/4 + 1/√2, 1 + 1/√2)

The equation of tangent is,

y – y_{1} = m (x – x_{1})

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

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

### Question 5. Find the equation of the tangent and the normal to the following curve at the indicated points.

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

**Solution:**

We have,

x = θ + sin θ and y = 1 + cos θ

dx/dθ = 1 + cos θ and dy/dθ = -sinθ

=

Slope of tangent, m=

= -1/(1 + 0)

= -1

Given, (x_{1}, y_{1}) = (π/2 + sin π/2, 1 + cos π/2) = (π/2 + 1, 1)

The equation of tangent is,

y – y_{1} = m (x – x_{1})

y – 1 = -1 (x – π/2 – 1)

2y – 2 = – 2x + π + 2

x + 2y – π – 4 = 0

The equation of normal is,

y – y_{1} = -1/m (x – x_{1})

y – 1 = 1 (x – π/2 -1)

2y – 2 = 2x – π – 2

2x – 2y = π

### (ii) at t = 1/2

**Solution:**

We have,

dx/dt =

=

dy/dt =

=

Slope of tangent, m=

Given, (x_{1}, y_{1}) =

The equation of tangent is,

y – y_{1} = m (x – x_{1})

80y – 16a = 65x – 26a

65x – 80y – 10a = 0

13x – 16y – 2a = 0

The equation of normal is,

y – y_{1} = -1/m (x – x_{1})

65y – 13a = – 80x + 32a

80x + 65y – 45a = 0

16x + 13y – 9a = 0

### (iii) x = at^{2}, y = 2at at t = 1

**Solution:**

We have,

x = at^{2}, y = 2at

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

Slope of tangent, m= = 1

Given, (x_{1} , y_{1}) = (a, 2a)

The equation of tangent is,

y – y_{1} = m (x – x_{1})

y – 2a = 1 (x – a)

y – 2a = x – a

x – y + a = 0

Equation of normal:

y – y_{1} = -1/m (x – x_{1})

y – 2a = – 1 (x – a)

y – 2a = – x + a

x + y = 3a

### (iv) x = a sec t, y = b tan t at t

**Solution:**

We have,

x = a sec t, y = b tan t

dx/dt = a sect tant and dy/dt = b sec^{2}t

Slope of tangent, m =

Given (x_{1}, y_{1}) = (a sec t, b tan t)

The equation of tangent is,

y – y_{1} = m (x – x_{1})

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

ay sin t cos t – ab sin^{2} t = bx cos t – ab

bx cos t – ay sin t cos t – ab (1 – sin^{2} t) = 0

bx cos t – ay sin t cos t = ab cos^{2} t

On dividing by cos^{2} t, we get

bx sec t – ay tan t = ab

The equation of normal is,

y – y_{1} = -1/m (x – x_{1})

y – b tant = -a/b sint(x – asect)

ycost − bsint = − a/bsint(xcost − a)

by cos t – b^{2} sin t = – ax sin t cos t + a^{2} sin t

ax sin t cos t + by cos t = (a^{2} + b^{2}) sin t

On dividing both sides by sin t, we get

ax cos t + by cot t = a^{2 }+ b^{2}

### (v) x = a(θ + sin θ), y = a(1 − cos θ) at θ

**Solution:**

We have,

x = a(θ + sin θ), y = a(1 − cos θ)

dx/dθ = a(1 + cosθ) and dy/dθ = asinθ

=

=

=

= tan θ/2 . . . . (1)

Slope of tangent, m=

Given (x_{1}, y_{1}) = [a(θ + sin θ), a(1 − cos θ)]

The equation of tangent is,

y – y_{1} = m (x – x_{1})

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

y − a(2sin^{2}θ/2) = xtanθ/2 − aθtanθ/2 − atanθ/2sinθ

y − 2asin^{2}θ/2 =(x − aθ)tan θ/2 − 2asin^{2}θ/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(2sin^{2}θ/2)] = −x + aθ + asinθ

tanθ/2[y − a{2(1 − cos^{2}θ/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

**Solution:**

We have,

x = 3 cos θ − cos^{3} θ, y = 3 sin θ − sin^{3} θ

dx/dθ = -3sin θ + 3 cos^{2}θ sin θ and dy/dθ = 3 cos θ – 3 sin^{2}θ cos θ

=

=

= cos^{3} θ/ -sin^{3} θ

= tan^{3} θ

So the equation of the tangent at θ is,

y – 3 sin θ + sin^{3 }θ = -tan^{3 }θ (x – 3 cos θ + cos^{3} θ)

4 (y cos^{3 }θ – x sin^{3 }θ) = 3 sin 4θ

So the equation of normal at θ is,

y – 3 sin θ + sin^{3} θ= (1/tan^{3} θ) (x – 3 cos θ + cos^{3} θ)

sin^{3 }θ – x cos^{3} θ = 3sin^{4 }θ – sin^{6} θ – 3cos^{4 }θ + cos^{6 }θ

### Question 6. Find the equation of the normal to the curve x^{2} + 2y^{2} − 4x − 6y + 8 = 0 at the point whose abscissa is 2?

**Solution:**

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

x^{2} + 2y^{2} − 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

When x = 2, we get

4 + 2y^{2} – 8 – 6y + 8 = 0

2y^{2} – 6y + 4 = 0

y^{2} – 3y + 2 = 0

y = 2 or y = 1

m (tangent) at x = 2 is 0

Normal is perpendicular to tangent so, m_{1}m_{2} = –1

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

The equation of normal is given by y – y_{1} = m (normal) (x – x_{1})

x = 2

### Question 7. Find the equation of the normal to the curve ay^{2} = x^{3} at the point (am^{2}, am^{3}).

**Solution:**

We have,

ay

^{2}= x^{3}On differentiating both sides w.r.t. x, we get

2aydy/dx = 3x

^{2}dy/dx = 3x

^{2}/2aySlope of tangent =

Given (x

_{1}, y_{1}) = (am^{2}, am^{3})The equation of normal is,

y – y

_{1}= -1/m (x – x_{1})y – a m

^{3}= -2m/3 (x – am^{2})3my – 3am

^{4}= – 2x + 2am^{2}2x + 3my – am

^{2}(2 + 3m^{2}) = 0

### Question 8. The equation of the tangent at (2, 3) on the curve y^{2} = ax^{3} + b is y = 4x − 5. Find the values of a and b.

**Solution:**

We have,

y

^{2}= ax^{3}+ bOn differentiating both sides w.r.t. x, we get

2y dy/dx = 3ax

^{2}dy/dx = 3ax

^{2}/2ySlope of tangent, m =

The equation of tangent is given by y – y

_{1}= m (tangent) (x – x_{1})Now compare the slope of a tangent with the given equation

2a = 4

a = 2

Now (2, 3) lies on the curve, these points must satisfy

32 = 2 (23) + b

b = – 7

### Question 9. Find the equation of the tangent line to the curve y = x^{2} + 4x − 16 which is parallel to the line 3x − y + 1 = 0.

**Solution:**

We have,

y = x^{2} + 4x − 16

Let (a, b) be the point of intersection of both the curve and the tangent.

Since (a, b) lies on curve, we get

b = a^{2} + 4a − 16

Now, x^{2} + 4x − 16

dy/dx = 2x + 4

Slope of tangent =

Given that the tangent is parallel to the line we have,

Slope of tangent = Slope of the given line

=> 2a + 4 = 3

=> 2 a = -1

=> a = -1/2

From eq(1), we get

b = 1/4 – 2 – 16 = -71/4

Now, slope of tangent, m = 3

(a, b) = (-1/2, -71/4)

The equation of tangent is,

y – y_{1} = m (x – x_{1})

y + 71/4 = 3 (x + 1/2)

4y + 71 = 12x + 6

12x – 4y – 65 = 0

### Question 10. Find an equation of normal line to the curve y = x^{3} + 2x + 6 which is parallel to the line x+ 14y + 4 = 0.

**Solution:**

We have,

y = x^{3} + 2x + 6

Let (a, b) be a point on the curve where we need to find the normal.

Slope of the given line = -1/14

Since the point lies on the curve, we get

b = a^{3} + 2a + 6

Now, y = x^{3 }+ 2x + 6

dy/dx = 3 x^{2 }+ 2

Slope of the tangent, m =

Slope of the normal =

Given that, slope of the normal = slope of the given line, we have

3a^{2} + 2 = 14

3a^{2} = 12

a^{2} = 4

a = ±2

So, b = 18 or -6.

And slope of the normal = -1/14

When a = 2 and b = 18, we have

y – y_{1 }= m (x – x_{1})

y – 18 = -1/14 (x – 2)

14y – 252 = -x + 2

x + 14y – 254 = 0

When a = -2 and b = -6, we have

y – y_{1} = m (x – x_{1})

y + 6 = -1/14 (x + 2)

14y + 84 = -x – 2

x + 14y + 86 = 0

### Question 11. Determine the equation(s) of tangent (s) line to the curve y = 4x^{3} − 3x + 5 which are perpendicular to the line 9y + x + 3 = 0.

**Solution:**

Let (a, b) be a point on the curve where we need to find the tangent(s).

Slope of the given line = -1/9

Since, tangent is perpendicular to the given line,

Slope of the tangent = = 9

Hence, b = 4 a

^{3}– 3 a + 5Now, y = 4 x

^{3}– 3x + 5dy/dx = 12 x

^{2}– 3Slope of the tangent =

Given that, slope of the tangent = slope of the perpendicular line

12a

^{2}– 3 = 912a

^{2}= 12a

^{2}= 1a = ±1

So, b = 6 or 4.

Thus, slope of tangent = 9.

When a = 1 and b = 6, we have

y – y

_{1}= m (x – x_{1})y – 6 = 9 (x – 1)

y – 6 = 9x – 9

9x – y – 3 = 0

When a = -1 and b = 4, we have

y – y

_{1}= m (x – x_{1})y – 4 = 9 (x + 1)

y – 4 = 9x + 9

9x – y + 13 = 0

### Question 12. Find the equation of a normal to the curve y = x log_{e} x which is parallel to the line 2x − 2y + 3 = 0.

**Solution:**

Slope of the given line is 1.

Let (a, b) be the point where the tangent is drawn to the curve.

Hence, b = a log_{e} a . . . . (1)

Now, y = x log_{e} x

dy/dx = x × 1/x + log_{e} x(1) = 1 + log_{e} x_{1}

Slope of tangent = 1 + log a

Slope of normal =

Given that, slope of normal = slope of the given line.

=> -1 = 1 + log a

=> – 2 = log a

=> a = e^{-2}

From (1), we have

Now, b = e^{-2} (-2) = -2 e^{-2}

Given, (x_{1}, y_{1}) = (e^{-2}, -2 e^{-2})

The equation of normal is,

y + 2/e^{2} = 1(x – 1/e^{2})

y + 2/e^{2} = x – 1/e^{2}

x – y = 3/e^{2}

### Question 13. Find the equation of the tangent line to the curve y = x^{2} − 2x + 7

### (i) which is parallel to the line 2x − y + 9 = 0?

**Solution:**

We have, y = x

^{2}− 2x + 7On differentiating both sides, we get

dy/dx = 2x – 2

The equation of the line is 2x – y + 9 = 0

So the slope of line is 2.

According to the question,

=> 2x – 2 = 2

=> 2x = 4

=> x = 2

=> y = 2

^{2}− 2(2) + 7 = 4 – 4 + 7 = 7As (x

_{1}, y_{1}) is (2, 7), the equation of tangent is,y – 7 = 2(x – 2)

y – 7 = 2x – 4

y – 2x – 3 = 0

### (ii) which is perpendicular to the line 5y − 15x = 13.

**Solution:**

We have, y = x

^{2}− 2x + 7On differentiating both sides, we get

dy/dx = 2x – 2

The equation of the line is 5y − 15x = 13.

=> y = 3x + 13/5

So the slope of line is 3.

According to the question,

=> 2x – 2= -1/3

=> 6x – 6 = -1

=> x = 5/6

And y = 217/36.

As (x

_{1}, y_{1}) is (5/6, 217/36), the equation of tangent is,y – y

_{1}= m (x – x_{1})y – 217/36 = (-1/3) (x – 5/6)

36y -217 = -12x + 10

36y + 12x – 227 = 0

### Question 14. Find the equations of all lines having slope 2 and that are tangent to the curve y = 1/x – 3, x ≠ 3.

**Solution:**

Let (a , b) be the point where the tangent is drawn to this curve.

Since, the point lies on the curve, hence b = 1/(a – 3)

Slope of tangent, m =

Slope of the tangent} = 2

(a – 3)^{2} = – 2

a – 3 = √-2, which does not exist because 2 is negative.

So, there does not exist any such tangent.

### Question 15. Find the equations of all lines of slope zero and that are tangent to the curve .

**Solution:**

Slope of the given tangent is 0.

Let (a, b) be a point where the tangent is drawn to the curve.

Since, the point lies on the curve, hence b = . . . . (1)

Slope of tangent =

Given that, slope of tangent = slope of the given line,

=> -2 a + 2 = 0

=> 2a = 2

=> a = 1

From (1), we get

Now, b =

(a, b) = (1, 1/2)

The equation of tangent is,

y – y_{1} = m (x – x_{1})

y – 1/2 = 0 (x – 1)

y = 1/2

### Question 16. Find the equation of the tangent to the curve which is parallel to the 4x − 2y + 5 = 0.

**Solution:**

We have,

Let (a, b) be the point where the tangent is drawn to the curve y =

On differentiating both sides, we get

Slope of tangent at (a, b) =

Slope of line 4x − 2y + 5 = 0 is 2.

Given that, slope of tangent = slope of the given line

9 = 16 (3a – 2)

9/16 = 3a – 2

3a = 9/16 + 2

a = 41/48

Now, b =

Therefore, (a, b) = (41/48, 3/4)

The equation of tangent is,

y – y_{1} = m (x – x_{1})

y – 3/4 = 2 (x – 41/48)

(4y – 3)/4 = 2 (48x – 41)/48

24y – 18 = 48x – 41

48x – 24y – 23 = 0

### Question 17. Find the equation of the tangent to the curve x^{2} + 3y − 3 = 0, which is parallel to the line y= 4x − 5.

**Solution:**

Suppose (a, b) be the required point.

We can find the slope of the given line by differentiating the equation w.r.t x,

So, slope of the line = 4

Since (a, b) lies on the curve, we get a

^{2}+ 3b − 3 = 0 . . . . (1)Now,

2x + 3dy/dx = 0

dy/dx = -2x/3

Slope of tangent, m=

Given that tangent is parallel to the line, So we get,

Slope of tangent, m = slope of the given line

=> -2a/3 = 4

=> a = -6

From (1), we get

=> 36 + 3b – 3 = 0

=> 3b = – 33

=> b = – 11

(a, b) = (-6, -11)

The equation of tangent is,

y – y

_{1}= m (x – x_{1})y + 11 = 4 (x + 6)

y + 11 = 4x + 24

4x – y + 13 = 0

### Question 18. Prove that touches the straight line for all n ∈ N, at the point (a, b) ?

**Solution:**

We have,

On differentiating both sides, we get

Slope of tangent =

The equation of tangent is,

y – b = -b/a (x – a)

ya – ab = – xb + ab

xb + ya = 2ab

So, the given line touches the given curve at the given point.

**Hence proved.**

### Question 19. Find the equation of the tangent to the curve x = sin 3t, y = cos 2t at t = π/4.

**Solution:**

We have,

x = sin 3t, y = cos 2t

dx/dt = 3 cos3t and dy/dt = -2sin2t

Slope of tangent, m=

x_{1} = sin 3π/4 = 1/√2 and y_{1} = cos π/2 = 0

So, (x_{1}, y_{1}) = (1/√2, 0)

The equation of tangent is,

y – y_{1} = m (x – x_{1})

3y = 2√2 x – 2

2√2 x – 3y – 2 = 0

### Question 20. At what points will be tangents to the curve y = 2x^{3} − 15x^{2} + 36x − 21 be parallel to x-axis? Also, find the equations of the tangents to the curve at these points?

**Solution:**

We have,

y = 2x

^{3}− 15x^{2}+ 36x − 21The slope of x – axis is 0.

Let (a, b) be the required point.

Since (a, b) lies on the curve, we get

b = 2a

^{3}− 15a^{2}+ 36a − 21 . . . . (1)Also, we have

dy/dx = 6 x

^{2}– 30x + 36Slope of tangent at (a, b) =

Given that the slope of the tangent = slope of the x-axis, we have

=> 6a

^{2}– 30a + 36 = 0=> a

^{2}– 5a + 6 = 0=> (a – 2) (a – 3) = 0

=> a = 2 or a= 3

=> b = 7 or 6

When a = 2 and b = 7, the equation is,

y – y

_{1}= m (x – x_{1})y – 7 = 0 (x – 2)

y = 7

When a = 3 and b = 6, the equation is,

y – y

_{1}= m (x – x_{1})y – 6 = 0 (x – 3)

y = 6

### Question 21. Find the equation of the tangents to the curve 3x^{2} – y^{2} = 8, which passes through the point (4/3, 0).

**Solution:**

We have,

3x

^{2}– y^{2}= 8 . . . . (1)On differentiating both sides w.r.t x, we get

6x – 2y dy/dx = 0

2y dy/dx = 6x

dy/dx = 6x/2y

dy/dx = 3x/y

Let tangent at (h, k) pass through (4/3, 0). Since, (h, k) lies on (1), we get

3 h

^{2}– k^{2}= 8 . . . (ii)Slope of tangent at (h, k) = 3h/k

The equation of tangent at (h, k) is given by,

y – k = 3h/k (x – h)

Also,

=> 0 – k = (3h/k) (4/3 – h)

=> -k = 4h/k – 3h

^{2}/k=> – k

^{2}= 4h – 3h^{2}=> 8 – 3 h

^{2}= 4h – 3 h^{2}=> 8 = 4h

=> h = 2

Also we get,

=> 12 – k

^{2}= 8=> k

^{2}= 4=> k = ±2

So, the points on curve (i) at which tangents pass through (4/3, 0) are (2, ±2).

When h = 2 and k = 2, the equation is,

y – 2 = (6/2) (x – 2)

3x – y – 4 = 0

When h = 2 and k = –2, the equation is,

y + 2 = (6/-2) (x – 2)

3x + y – 4 = 0

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