Tangents and Normals

The normal is perpendicular to the tangent, so its gradient is the negative reciprocal of the tangent's gradient.

1. The tangent at P has gradient 3
2. The normal is perpendicular to it: gradient =-\dfrac{1}{3} (flip it and change the sign)
3. It passes through (-2,\ 5): \;y-5=-\dfrac13(x+2)
4. Answer: y-5=-\frac{1}{3}(x+2)

Substitute x=2 into \frac{dy}{dx} to get the gradient of the tangent first.

1. \dfrac{dy}{dx}=3x^2-5
2. At x=2: gradient of the tangent =3(4)-5=7
3. The normal's gradient is the negative reciprocal of 7
4. Answer: -\dfrac{1}{7}

Rearrange the normal's equation into the form y=mx+c to read off its gradient; the curve's gradient is the gradient of the tangent.

1. x-3y=6 gives 3y=x-6, so y=\dfrac13x-2
2. The normal has gradient \dfrac13
3. The tangent is perpendicular to the normal: gradient =-3
4. The gradient of the curve at P is the gradient of the tangent
5. Answer: -3

You need a point and a gradient: find y when x=2, then substitute x=2 into \frac{dy}{dx}.

1. When x=2: y=2(8)-5(2)+1=7, so the point is (2,\ 7)
2. \dfrac{dy}{dx}=6x^2-5
3. When x=2: gradient =6(4)-5=19
4. y-7=19(x-2)
5. y=19x-31

At a turning point the tangent is horizontal, so the normal is vertical.

1. \dfrac{dy}{dx}=4x+12=0 at the turning point, so x=-3
2. y=2(9)+12(-3)+5=-13, so M is (-3,\ -13)
3. The tangent at M is horizontal (y=-13), so the normal is the vertical line through M
4. Answer: x=-3

Expand the brackets before you differentiate.

1. When x=-1: y=(-3)(2)=-6, so the point is (-1,\ -6)
2. y=2x^2+5x-3
3. \dfrac{dy}{dx}=4x+5
4. When x=-1: gradient =-4+5=1
5. y+6=1(x+1)
6. Answer: y=x-5

The normal is perpendicular to the tangent: find the tangent's gradient, then flip it and change its sign.

1. \dfrac{dy}{dx}=3-2x
2. At x=4: gradient of the tangent =3-8=-5
3. Gradient of the normal is the negative reciprocal: \dfrac15
4. y-1=\dfrac15(x-4)
5. Multiply by 5: 5y-5=x-4
6. x-5y=-1

First find the x-coordinate of A by solving x^3-8=0.

1. At A, y=0: \;x^3=8, so x=2 and A is (2,\ 0)
2. \dfrac{dy}{dx}=3x^2
3. When x=2: gradient =3(4)=12
4. y-0=12(x-2)
5. Answer: y=12x-24

Parallel lines have the same gradient: find the gradient of 2x+y=7, then find where \frac{dy}{dx} equals it.

1. 2x+y=7 gives y=-2x+7, so its gradient is -2
2. \dfrac{dy}{dx}=2x-6
3. 2x-6=-2, so x=2
4. y=4-12+4=-4
5. Answer: P(2,\ -4)

Split the fraction into x^2-\frac{4}{x}, then write \frac{4}{x} as 4x^{-1} before you differentiate.

1. When x=2: y=\dfrac{8-4}{2}=2, so the point is (2,\ 2)
2. y=x^2-4x^{-1}
3. \dfrac{dy}{dx}=2x+4x^{-2}=2x+\dfrac{4}{x^2}
4. When x=2: gradient of the tangent =4+1=5
5. Gradient of the normal =-\dfrac15
6. y-2=-\dfrac15(x-2)
7. Multiply by 5: 5y-10=-x+2
8. Answer: x+5y=12

Find the gradient of the tangent at A; the tangent at B has the negative reciprocal of that gradient.

1. \dfrac{dy}{dx}=2x-3
2. At A: gradient =2-3=-1
3. Perpendicular gradient: -1\times m=-1, so m=1
4. At B: 2x-3=1
5. Answer: x=2

Write \frac{4}{x^2} as 4x^{-2}, find the tangent at x=2, then put y=0 into its equation.

1. When x=2: y=6+\dfrac44=7, so the point is (2,\ 7)
2. y=3x+4x^{-2}, so \dfrac{dy}{dx}=3-8x^{-3}=3-\dfrac{8}{x^3}
3. When x=2: gradient =3-1=2
4. Tangent: y-7=2(x-2), so y=2x+3
5. At T, y=0: 2x+3=0, so x=-\dfrac32
6. Answer: T\left(-\frac32,\ 0\right)

Find the point and the gradient of the tangent at x=3, take the negative reciprocal, then put x=0 into the normal's equation.

1. When x=3: y=27-27+4=4, so the point is (3,\ 4)
2. \dfrac{dy}{dx}=3x^2-6x
3. When x=3: gradient of the tangent =27-18=9
4. Gradient of the normal =-\dfrac19
5. Normal: y-4=-\dfrac19(x-3)
6. At N, x=0: y-4=\dfrac39=\dfrac13, so y=\dfrac{13}{3}
7. Answer: N\left(0,\ \frac{13}{3}\right)

Use the normal's gradient to find the gradient of the tangent at x=2, then set \frac{dy}{dx} at x=2 equal to it.

1. The normal has gradient \dfrac12, so the tangent has gradient -2
2. y=x^2+kx^{-1}, so \dfrac{dy}{dx}=2x-kx^{-2}=2x-\dfrac{k}{x^2}
3. When x=2: \;4-\dfrac{k}{4}=-2
4. \dfrac{k}{4}=6
5. Answer: k=24

The curve and the tangent have the same gradient at x=2, which gives b; they also share the point where x=2.

1. \dfrac{dy}{dx}=2x+b
2. At x=2 the gradient is 3 (the gradient of the tangent): 4+b=3, so b=-1
3. The point where x=2 is on the tangent: y=3(2)-1=5
4. It is also on the curve: 4+2b+c=5
5. 4-2+c=5
6. Answer: c=3

For (b), use \frac{dy}{dx} to find where the gradient is 8, then find the tangent there and solve it simultaneously with your tangent from (a).

Part (a)

1. When x=1: y=1+2-4=-1, so A is (1,\ -1)
2. \dfrac{dy}{dx}=2x+2
3. When x=1: gradient =2+2=4
4. y-(-1)=4(x-1)
5. y+1=4x-4
6. y=4x-5

Part (b)

1. At B the gradient is 8: \;2x+2=8, so x=3
2. When x=3: y=9+6-4=11, so B is (3,\ 11)
3. Tangent at B: \;y-11=8(x-3), so y=8x-13
4. At R the two tangents meet: \;4x-5=8x-13
5. 8=4x, so x=2
6. y=4(2)-5=3
7. (Check in the other tangent: 8(2)-13=3 ✓)
8. R(2,\ 3)

At a turning point \frac{dy}{dx}=0, so the tangent there is a horizontal line. Answer part (a) before attempting part (b).

Part (a)

1. \dfrac{dy}{dx}=2x-2=0 at the turning point, so x=1
2. y=1-2+7=6, so P is (1,\ 6)
3. The tangent at a turning point is horizontal
4. Answer: y=6

Part (b)

1. When x=2: y=4-4+7=7, so Q is (2,\ 7)
2. When x=2: gradient =2(2)-2=2
3. Gradient of the normal =-\dfrac12
4. Normal: y-7=-\dfrac12(x-2)
5. At R, y=6: -1=-\dfrac12(x-2)
6. x-2=2, so x=4
7. Answer: R(4,\ 6)

Part (c)

1. PR lies along y=6 from x=1 to x=4: base =3
2. Q is 1 unit above y=6: height =1
3. Area =\dfrac12\times3\times1
4. Answer: \dfrac32

Factorise to find A, then solve the equation of the normal simultaneously with the curve; you already know one of the solutions. Answer part (a) before attempting part (b).

Part (a)

1. x^2-5x+6=(x-2)(x-3)=0, so A is (3,\ 0)
2. \dfrac{dy}{dx}=2x-5; at x=3 the gradient is 1
3. Gradient of the normal =-1
4. Normal: y-0=-1(x-3), so y=3-x
5. Meets the curve: x^2-5x+6=3-x
6. x^2-4x+3=0, so (x-1)(x-3)=0
7. x=3 is A, so at B: x=1 and y=3-1=2
8. Answer: B(1,\ 2)

Part (b)

1. Horizontal distance =3-1=2; vertical distance =2-0=2
2. AB^2=2^2+2^2=8
3. AB=\sqrt8=\sqrt4\times\sqrt2
4. Answer: 2\sqrt2

For (b), substitute your tangent from (a) into the equation of the circle: a tangent meets the circle at only one point. Answer part (a) before attempting part (b).

Part (a)

1. When x=3: y=9-12-1=-4, so P is (3,\ -4)
2. \dfrac{dy}{dx}=2x-4
3. When x=3: gradient =6-4=2
4. y+4=2(x-3)
5. Answer: y=2x-10

Part (b)

1. Substitute y=2x-10 into x^2+y^2=20
2. x^2+(2x-10)^2=20
3. x^2+4x^2-40x+100=20
4. 5x^2-40x+80=0, so x^2-8x+16=0
5. (x-4)^2=0, so x=4 (one repeated root, as it is a tangent)
6. y=2(4)-10=-2
7. Answer: (4,\ -2)

Call the x-coordinate of a point of contact p: write the tangent's gradient in terms of p and use the fact that the tangent passes through T. Answer part (a) before attempting part (b).

Part (a)

1. Let a point of contact be (p,\ p^2-2p+5)
2. The gradient there is 2p-2
3. The tangent through T(1,\ 0) with this gradient is y=(2p-2)(x-1)
4. The point of contact is on this line, so put x=p:
5. p^2-2p+5=(2p-2)(p-1)
6. p^2-2p+5=2p^2-4p+2
7. p^2-2p-3=0, so (p-3)(p+1)=0
8. Answer: x=-1 and x=3

Part (b)

1. At x=3: y=9-6+5=8 and the gradient is 2(3)-2=4
2. y-8=4(x-3)
3. Answer: y=4x-4