# Class 12 RD Sharma Solutions – Chapter 16 Tangents and Normals – Exercise 16.2 | Set 1

### 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}/4x + 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}θbx 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 θax sin θ – by cos θ = (a

^{2}– b^{2}) sin θ cos θOn 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 – abbx cos θ – ay sin θ cos θ = ab (1 – sin

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

^{2}θOn dividing by ab cos

^{2}θ, we getx/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 θax sin θ + by = (a

^{2}+ b^{2}) tan θOn 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}= 4axOn 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 – am

^{2}x – my + a = 0The equation of normal is,

y – y

_{1}= -1/m (x – x_{1})m

^{3}y – 2a m^{2}= – m^{2}x + am

^{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}θGiven, (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}θx cos

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

^{3}θ + y sin^{3}θ = cThe 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 2θsin

^{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 + ctx + y t

^{2}= 2ctThe 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 getx

_{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 getThe 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 getThe 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}= 4axOn differentiating both sides w.r.t. x, we get

2y dy/dx = 4a

dy/dx = 2a/y

At (x

_{1}, y_{1}), we haveSlope 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 = 2atdx/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}tSlope 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 – abbx cos t – ay sin t cos t – ab (1 – sin

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

^{2}tOn dividing by cos

^{2}t, we getbx 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 tax sin t cos t + by cos t = (a

^{2}+ b^{2}) sin tOn 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/2sinθy − 2asin

^{2}θ/2 =(x − aθ)tan θ/2 − 2asin^{2}θ/2y = (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 + asinθtanθ/2[y − a{2(1 − cos

^{2}θ/2)}] = −x + aθ + asinθtan 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}θdx/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}θSo 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 4θSo 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 = 02y

^{2}– 6y + 4 = 0y

^{2}– 3y + 2 = 0y = 2 or y = 1

m (tangent) at x = 2 is 0

Normal is perpendicular to tangent so, m

_{1}m_{2}= –1m (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

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