Tangents and Normals

MATTangents & Normals

At a point on a curve $y=f(x)$ , the derivative $\dfrac{dy}{dx}$ gives the slope of the tangent line; the normal is perpendicular to it. ISC problems ask for tangent/normal equations at a point, tangents with a given slope or parallel/perpendicular to a line, points with horizontal or vertical tangents, and the angle of intersection of two curves.

Mastery of point-slope form plus the perpendicular-slope rule makes this a dependable scoring topic.

Slope of tangent and normal

m_{t}=\left.\dfrac{dy}{dx}\right|_{(x_1,y_1)},\qquad m_{n}=-\dfrac{1}{m_{t}}

m_t

is the tangent slope at

(x_1,y_1)

;

m_n

is the normal slope, valid when

m_t\ne0

Equation of the tangent

y-y_{1}=m_{t}(x-x_{1})

(x_1,y_1)

is the point of contact and

m_t

the tangent slope there.

Equation of the normal

y-y_{1}=-\dfrac{1}{m_{t}}(x-x_{1})

(x_1,y_1)

is the point of contact; for

m_t=0

the normal is the vertical line

x=x_1

Angle of intersection of two curves

\tan\theta=\left|\dfrac{m_{1}-m_{2}}{1+m_{1}m_{2}}\right|

m_1,m_2

are the tangent slopes of the two curves at their common point;

\theta

is the acute angle between them.

Horizontal tangent (parallel to the $x$ -axis) occurs where $\dfrac{dy}{dx}=0$ ; vertical tangent where $\dfrac{dy}{dx}$ is undefined.
A tangent parallel to a given line shares its slope; a tangent perpendicular to it has the negative reciprocal slope.
For a parallel-to-line condition, set $\dfrac{dy}{dx}$ equal to the line's slope, solve for the point, then write the line.
Tangent and normal at the same point are always perpendicular, so $m_t\cdot m_n=-1$ .
Two curves cut orthogonally when $m_1 m_2=-1$ ; they touch (angle $0$ ) when $m_1=m_2$ .
For implicit curves, use implicit differentiation to get $\dfrac{dy}{dx}$ before substituting the point.
Always evaluate the slope AT the given point — $\dfrac{dy}{dx}$ is generally a function of $x$ (and $y$ ).

Where the marks go

Using the normal slope as $-m_t$ instead of $-\dfrac{1}{m_t}$ (negative reciprocal, not just negative).
Forgetting to substitute the point's coordinates into $\dfrac{dy}{dx}$ , leaving a symbolic slope.
Mishandling $m_t=0$ (normal is vertical $x=x_1$ ) or $m_t$ undefined (tangent is vertical).
Dropping the absolute value in the angle-of-intersection formula, giving an obtuse or negative angle.

How the board asks it

Numericalequation of the tangent and normal at a point
Find the equations of the tangent and the normal to the curve $y = x^3 - 2x + 7$ at the point $(1, 6)$ .
Numericaltangent parallel or perpendicular to a given line
Find the equation of the tangent to the curve $y = x^2 - 2x + 3$ that is parallel to the line $2x - y + 9 = 0$ .
Numericalhorizontal or vertical tangent points where $\frac{dy}{dx}=0$
Find the points on the curve $y = x^3 - 3x^2 - 9x + 7$ at which the tangent is parallel to the $x$ -axis.
Numericalangle of intersection of two curves
Find the angle of intersection of the curves $y^2 = 4x$ and $x^2 = 4y$ .
Derive / provetwo curves cut orthogonally when $m_1 m_2 = -1$
Show that the curves $x^2 + y^2 = a^2$ and $xy = c^2$ cut each other orthogonally.
Numericalimplicit differentiation before substituting the point
Find the equation of the normal to the curve $x^2 + 3xy + y^2 = 5$ at the point $(1, 1)$ .

Practise this topic

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Written for Sublevo. Question text quoted anywhere in these notes is the Council’s and carries its year and paper; the board’s own diagrams are not reproduced.