Geometrical Applications of Derivatives

Q: How do you find the equation of the tangent and normal to a curve?

Find the slope m = f'(x₁) at the point (x₁, y₁). The tangent is y - y₁ = m(x - x₁) and the normal is y - y₁ = -1/m(x - x₁). If m = 0 the tangent is horizontal and the normal is the vertical line x = x₁.

Q: What is the geometrical meaning of dy/dx?

dy/dx at a point is the slope of the tangent to the curve there, equal to tan ψ where ψ is the angle the tangent makes with the positive x-axis. It is the limit of the slopes of chords through the point.

Q: How do you find the tangent from an external point to a curve?

Take the unknown point of contact (h, f(h)). The tangent there has slope f'(h) and must pass through the given point (a, b), so solve f'(h) = (f(h) - b)/(h - a). Each real root gives one tangent.

Q: What are subtangent and subnormal?

They are the projections on the x-axis of the tangent and normal segments from the point to the x-axis. For a point (h, k) with slope m, the subtangent is |k/m| and the subnormal is |km|.

Q: How do you find the angle between two curves?

Find their point of intersection, compute the slopes m₁ and m₂ of the two tangents there, and use tan θ = |(m₁ - m₂)/(1 + m₁m₂)|. The angle is the acute angle between the tangents.

MathsApplication Of DerivativesFor JEE aspirants

Geometrical applications of derivatives use one fact: $\frac{d y}{d x}$ at a point is the slope of the tangent there. From it we get the equations of the tangent and the normal, tangents drawn from an outside point, the lengths of the tangent, normal, subtangent and subnormal, the angle at which two curves cut, and the shortest distance between two curves. These geometrical applications of derivatives are tested every year in JEE Main and JEE Advanced, often mixed with conics.

On this page1Tangent and normal2Special tangents3From an external point4Lengths5Angle between curves6Shortest distance7Examples

Key Formulas - Quick Reference

★ Must learnSlope of the tangent at $(x_{1}, y_{1})$ : $m = \frac{d y}{d x}_{(x_{1}, y_{1})} = f^{'} (x_{1}) = tan ψ$
★ Must learnTangent: $y - y_{1} = f^{'} (x_{1}) (x - x_{1})$ ; normal: $y - y_{1} = - \frac{1}{f ^{'} ( x _{1} )} (x - x_{1})$
$f^{'} (x_{1}) = 0$ : tangent $y = y_{1}$ , normal $x = x_{1}$ ; $f^{'} (x) \to \pm \infty$ : vertical tangent $x = x_{1}$
Parametric curve: $\frac{d y}{d x} = \frac{d y / d t}{d x / d t}$
Tangent from $(a, b)$ not on the curve: contact point $(h, f (h))$ with $f^{'} (h) = \frac{f ( h ) - b}{h - a}$
★ Must learnAt $P (h, k)$ with slope $m$ : tangent length $∣ k ∣ 1 + \frac{1}{m ^{2}}$ , normal length $∣ k ∣ 1 + m^{2}$ , subtangent $\frac{k}{m}$ , subnormal $∣ k m ∣$
★ Must learnAngle between curves: $tan θ = \frac{m _{1} - m _{2}}{1 + m _{1} m _{2}}$ ; orthogonal when $m_{1} m_{2} = - 1$
★ Must learnShortest distance between two non-intersecting curves is along their common normal

1. Tangent and Normal

Let $P$ be a point on the curve $y = f (x)$ and $Q$ a nearby point on it. The line $P Q$ is a secant. As $Q$ slides along the curve towards $P$ (through $Q_{1}, Q_{2}, \dots$ ), the secant $P Q$ turns and settles into a limiting position: the tangent at $P$ . The line through $P$ perpendicular to the tangent is the normal at $P$ .

1.1 Geometrical meaning of $\frac{d y}{d x}$

Take $P (x, f (x))$ and $Q (x + h, f (x + h))$ . The slope of the chord $P Q$ is $\frac{f ( x + h ) - f ( x )}{h}$ . As $Q \to P$ , $h \to 0$ , and

slope of tangent at P = h \to 0 lim \frac{f ( x + h ) - f ( x )}{h} = f^{'} (x) = \frac{d y}{d x} .

If the tangent at $P$ makes an angle $ψ$ with the positive $x$ -axis, then $\frac{d y}{d x} = tan ψ$ = slope of the tangent at $P$ . The normal, being perpendicular, has slope $- \frac{1}{tan ψ} = - \frac{d x}{d y}$ .

Figure 1: Tangent and normal to

y = e^{x}

P (0, 1)

. The slope is

\frac{d y}{d x}_{x = 0} = e^{0} = 1 = tan ψ

, so the tangent makes

ψ = 4 5^{\circ}

with the

x

-axis:

y = x + 1

. The normal is perpendicular to it (slope

- 1

y = - x + 1

1.2 Equations of tangent and normal

$\frac{d y}{d x}_{(x_{1}, y_{1})} = f^{'} (x_{1})$ is the slope of the tangent at $(x_{1}, y_{1})$ on $y = f (x)$ . So

Tangent: y - y_{1} = f^{'} (x_{1}) (x - x_{1}) (f^{'} (x_{1}) real)

Normal: y - y_{1} = - \frac{1}{f ^{'} ( x _{1} )} (x - x_{1}) (f^{'} (x_{1}) \neq = 0)

Slope at $(x_{1}, y_{1})$	Tangent	Normal
$f^{'} (x_{1}) = m \neq = 0$	$y - y_{1} = m (x - x_{1})$	$y - y_{1} = - \frac{1}{m} (x - x_{1})$
$f^{'} (x_{1}) = 0$ (top of a hump, bottom of a dip)	$y = y_{1}$ (horizontal)	$x = x_{1}$ (vertical)
$f^{'} (x) \to \pm \infty$ as $x \to x_{1}$ (e.g. $y = x^{1/3}$ at $O$ )	$x = x_{1}$ (vertical)	$y = y_{1}$ (horizontal)

Figure 2: Left:

y = x^{3} - 3 x

has

f^{'} (x) = 3 x^{2} - 3 = 0

x = \pm 1

, so the tangents there are horizontal (

y = 2

and

y = - 2

). Right:

y = x^{1/3}

has

f^{'} (x) = \frac{1}{3} x^{- 2/3} \to \infty

x \to 0

, so the tangent at the origin is vertical (

x = 0

) and the normal is

y = 0

1.3 Parametric and implicit curves

For a curve given by $x = x (t)$ , $y = y (t)$ , the slope is $\frac{d y}{d x} = \frac{d y / d t}{d x / d t}$ , evaluated at the parameter value of the point. For an implicit curve $F (x, y) = 0$ , differentiate both sides with respect to $x$ and solve for $\frac{d y}{d x}$ ; then substitute the coordinates of the point.

Exam Trick

Tangent to any second-degree curve in one line. For $a x^{2} + 2 h x y + b y^{2} + 2 g x + 2 f y + c = 0$ , the tangent at $(x_{1}, y_{1})$ is obtained by replacing $x^{2} \to x x_{1}$ , $y^{2} \to y y_{1}$ , $2 x y \to x y_{1} + x_{1} y$ , $2 x \to x + x_{1}$ , $2 y \to y + y_{1}$ . Example: tangent to $x^{2} + y^{2} = 25$ at $(3, 4)$ is $3 x + 4 y = 25$ ; tangent to $y^{2} = 4 a x$ at $(x_{1}, y_{1})$ is $y y_{1} = 2 a (x + x_{1})$ .

1.4 Tangents with a given slope

When the slope $m$ is given (for example "parallel to a line" or "perpendicular to a line"), solve $f^{'} (x) = m$ . Each root gives a point of contact, and each point of contact gives one tangent. A curve with two branches can have one such tangent on each branch.

Figure 3: Tangents to

y = \frac{1}{x - 1}

with slope

- 1

(parallel to

x + y = 0

). Solving

- \frac{1}{( x - 1 ) ^{2}} = - 1

gives

x = 0

2

: one tangent on each branch,

x + y + 1 = 0

(0, - 1)

and

x + y = 3

(2, 1)

Curves with a modulus. Before differentiating $∣ g (x) ∣$ or $∣ x ∣$ , find the sign of the expression inside near the point and remove the modulus. Only that local piece decides the tangent.

Figure 4:

y = ∣ x^{2} - ∣ x ∣∣

has corners at

x = - 1, 0, 1

, but near

x = - 2

it is simply

y = x^{2} + x

. Slope

2 x + 1 = - 3

, so the normal at

(- 2, 2)

has slope

\frac{1}{3}

3 y = x + 8

Key idea

Slope first:

m = f^{'} (x_{1})

. Then tangent

y - y_{1} = m (x - x_{1})

and normal with slope

- \frac{1}{m}

;

m = 0

m \to \infty

swap horizontal and vertical.

2. Tangents and Normals from an External Point

Suppose $P (a, b)$ does not lie on $y = f (x)$ , and we want the tangents to the curve that pass through $P$ . The point of contact $Q (h, f (h))$ is unknown. The tangent at $Q$ has slope $f^{'} (h)$ , and it must pass through $P$ , so its slope also equals the slope of $P Q$ :

f^{'} (h) = \frac{f ( h ) - b}{h - a} .

Solve this for $h$ . Each real root $h$ gives one tangent: $y - b = \frac{f ( h ) - b}{h - a} (x - a)$ . The number of real roots is the number of tangents from $P$ .

Figure 5: Tangents from the external point

O (0, 0)

y = (x + 1)^{3}

. For a contact point

Q (h, (h + 1)^{3})

the slope

3 (h + 1)^{2}

must equal the slope of

O Q

, giving

h = - 1

h = \frac{1}{2}

: two tangents,

y = 0

and

4 y = 27 x

Tangent AT a point

The point $(x_{1}, y_{1})$ is on the curve. Slope $= f^{'} (x_{1})$ directly; exactly one tangent (if $f$ is differentiable there).

Tangent FROM a point

The point $(a, b)$ is off the curve. Contact $(h, f (h))$ unknown; solve $f^{'} (h) = \frac{f ( h ) - b}{h - a}$ . There may be 0, 1, 2 or more tangents.

2.1 A line meeting a curve

If a line touches a curve, then solving the line and the curve together gives at least a repeated (double) root at the point of contact. This is often the fastest test for tangency: substitute the line into the curve and make the resulting equation have equal roots ( $D = 0$ for a quadratic).

The same idea finds where a tangent cuts the curve again: the contact point is a double root, and the remaining root follows from the sum (or product) of roots.

Figure 6: The tangent

y = 12 x - 16

P (2, 8)

cuts

y = x^{3}

again. The intersection equation

x^{3} - 12 x + 16 = 0

must have

x = 2

as a double root (contact), so by the sum of roots

2 + 2 + h = 0

h = - 4

Q (- 4, - 64)

2.2 Normals through an external point

For normals through $P (a, b)$ , use the normal slope $- \frac{1}{f ^{'} ( h )}$ at $Q (h, f (h))$ and equate it to the slope of $P Q$ . The number of real roots tells how many normals pass through $P$ . For the parabola $x^{2} = 4 y$ the equation is a cubic in $h$ , so at most three normals can pass through a point; from $(1, 2)$ only one does.

Figure 7: Normals from

P (1, 2)

x^{2} = 4 y

. At

Q (h, \frac{h ^{2}}{4})

the normal slope is

- \frac{2}{h}

; requiring it to pass through

P

gives

h^{3} = 8

, so only

h = 2

is real. The single normal is

x + y = 3

, meeting the curve at right angles at

Q (2, 1)

Figure 8: Choosing the method. A point on the curve gives the slope directly; a given slope gives the contact points from

f^{'} (x) = m

; an external point

(a, b)

needs an unknown contact point

(h, f (h))

. For a line and a curve, contact means a repeated root.

Quick Recall: tap to check

What equation gives the point of contact of a tangent drawn from

(a, b)

f^{'} (h) = \frac{f ( h ) - b}{h - a}

, with contact point

(h, f (h))

A line touches a curve. What happens when you solve them together?

The point of contact appears as a repeated root.

How many tangents can be drawn from the origin to

y = (x + 1)^{3}

Two:

y = 0

(contact at

(- 1, 0)

) and

4 y = 27 x

(contact at

(\frac{1}{2}, \frac{27}{8})

Slope of the normal at a point where

f^{'} (x_{1}) = 0

Undefined: the normal is the vertical line

x = x_{1}

Key idea

From an outside point the contact point is the unknown: write it as

(h, f (h))

and make the tangent pass through the given point.

3. Lengths of Tangent, Normal, Subtangent and Subnormal

Let $P (h, k)$ be a point on $y = f (x)$ with $m = \frac{d y}{d x}_{(h, k)}$ . Let the tangent at $P$ meet the $x$ -axis at $T$ , the normal at $P$ meet it at $N$ , and let $M (h, 0)$ be the foot of the ordinate.

Tangent: $y - k = m (x - h)$ . Put $y = 0$ : $T = (h - \frac{k}{m}, 0)$ .
Normal: $y - k = - \frac{1}{m} (x - h)$ . Put $y = 0$ : $N = (h + k m, 0)$ .
Use the distance formula for $P T$ , $P N$ , and differences of $x$ -coordinates for $T M$ , $M N$ .

Figure 9: At

P (h, k)

with slope

m

, the tangent meets the

x

-axis at

T (h - \frac{k}{m}, 0)

and the normal at

N (h + k m, 0)

;

M

is the foot of the ordinate.

P T

and

P N

are the lengths of the tangent and normal; their projections

T M

and

M N

are the subtangent and subnormal.

Name	Segment	Length
Length of tangent	$P T$	$∣ k ∣ 1 + \frac{1}{m ^{2}}$
Length of normal	$P N$	$∣ k ∣ 1 + m^{2}$
Subtangent	$T M$ (projection of $P T$ on the $x$ -axis)	$\frac{k}{m}$
Subnormal	$M N$ (projection of $P N$ on the $x$ -axis)	$∣ k m ∣$

Two quick consequences: $(subtangent) \times (subnormal) = k^{2}$ , the square of the ordinate; and the subnormal $∣ k m ∣ = y \frac{d y}{d x}$ is often the easiest to compute for implicit curves.

Subtangent

$T M = \frac{y}{d y / d x}$ . Constant for $y = b e^{x / a}$ (equals $∣ a ∣$ ).

Subnormal

$M N = y \frac{d y}{d x}$ . Constant for $y^{2} = 4 a x$ (equals $2 a$ ).

3.1 Intercepts of a tangent

Writing the tangent in intercept form $\frac{x}{A} + \frac{y}{B} = 1$ gives the intercepts $A$ , $B$ on the axes. Many problems ask to show that a sum, a product or an area built from them is constant. For $x + y = a$ the tangent at $(x_{1}, y_{1})$ is $\frac{x}{x _{1}} + \frac{y}{y _{1}} = a$ , with intercepts $a x_{1}$ and $a y_{1}$ .

Figure 10: Two tangents to

x + y = 2

(

a = 4

). At

P (0.64, 1.44)

the intercepts are

1.6

and

2.4

; at

(2.25, 0.25)

they are

3

and

1

. In both cases the sum is

4 = a

: the intercepts are

a x_{1}

and

a y_{1}

, which add to

a \cdot a

Exam Trick

Constant-property curves to remember (each is a one-line answer in MCQs):

$x y = c^{2}$ : tangent and axes form a triangle of area $2 c^{2}$ , and $P$ is the midpoint of the intercepted segment.
$x + y = a$ : sum of intercepts $= a$ .
$x^{2/3} + y^{2/3} = a^{2/3}$ : part of the tangent between the axes has length $a$ .
$y = b e^{x / a}$ : subtangent $= ∣ a ∣$ ; $y^{2} = 4 a x$ : subnormal $= 2 a$ .

Key idea

All four lengths come from the ordinate

k

and the slope

m

P

: tangent

∣ k ∣ 1 + 1/ m^{2}

, normal

∣ k ∣ 1 + m^{2}

, subtangent

∣ k / m ∣

, subnormal

∣ k m ∣

4. Angle Between Two Curves

The angle between two intersecting curves is the acute angle between their tangents (equivalently, their normals) at a point of intersection $(x_{1}, y_{1})$ . If the slopes of the tangents there are $m_{1}$ and $m_{2}$ ,

tan θ = \frac{m _{1} - m _{2}}{1 + m _{1} m _{2}} .

The angle is defined only where the curves actually meet: find (or prove the existence of) the point of intersection first.
If the curves meet at several points, find the angle at each point separately.
Orthogonal curves cut at right angles at every point of intersection: $m_{1} m_{2} = - 1$ there.
$m_{1} = m_{2}$ means $θ = 0$ : the curves touch (common tangent). If one tangent is vertical and the other has inclination $ψ$ , then $θ = ∣9 0^{\circ} - ψ ∣$ .

Figure 11:

y^{2} = 4 x

and

x^{2} = 4 y

meet at

(0, 0)

and

(4, 4)

. At the origin one tangent is the

y

-axis and the other the

x

-axis, so

θ = 9 0^{\circ}

. At

(4, 4)

the slopes are

\frac{1}{2}

and

2

tan θ = \frac{2 - 1/2}{1 + 1} = \frac{3}{4}

θ \approx 36. 9^{\circ}

. The curves are not orthogonal.

JEE Advanced

Orthogonal central conics. The curves $a x^{2} + b y^{2} = 1$ and $a^{'} x^{2} + b^{'} y^{2} = 1$ cut orthogonally if and only if

\frac{1}{a} - \frac{1}{b} = \frac{1}{a ^{'}} - \frac{1}{b ^{'}} .

Reason: at a common point $m_{1} m_{2} = \frac{a a ^{'} x ^{2}}{b b ^{'} y ^{2}} = - 1$ , i.e. $a a^{'} x^{2} + b b^{'} y^{2} = 0$ ; substituting $x^{2}$ and $y^{2}$ from the two equations gives the condition. For an ellipse and a hyperbola it says they are confocal. Example: $x^{2} - y^{2} = 5$ and $\frac{x ^{2}}{18} + \frac{y ^{2}}{8} = 1$ give $5 + 5 = 18 - 8 = 10$ , so they are orthogonal.

4.1 Touching curves and the number of roots

An equation $g (x) = h (x)$ has as many real roots as the graphs $y = g (x)$ and $y = h (x)$ have common points. The boundary case between "two roots" and "no root" is when the curves touch: equal heights and equal slopes at the same $x$ .

Figure 12: Roots of

p x^{2} = ln x

as intersections of

y = p x^{2}

and

y = ln x

. For

p \leq 0

there is exactly one. For

p > 0

the parabola misses (

p = 0.4

), cuts twice (

p = 0.1

) or touches: equal slopes

2 p x = \frac{1}{x}

and equal heights give

x = e

p = \frac{1}{2 e} \approx 0.184

Exam Trick

Touching = two equations. For $y = g (x)$ and $y = h (x)$ to touch at $x = x_{1}$ : $g (x_{1}) = h (x_{1})$ and $g^{'} (x_{1}) = h^{'} (x_{1})$ . Solve the pair for the parameter. This settles "exactly one root" and "common tangent" questions in two lines.

Quick Recall: tap to check

Angle between curves at a point where

m_{1} m_{2} = - 1

9 0^{\circ}

(they cut orthogonally there).

What does

m_{1} = m_{2}

at a common point mean?

The curves touch: they have a common tangent there (

θ = 0

y^{2} = 4 x

and

x^{2} = 4 y

: angle at the origin?

9 0^{\circ}

; one tangent is the

y

-axis, the other the

x

-axis.

Key idea

Angle between curves = acute angle between tangents at the common point:

tan θ = \frac{m _{1} - m _{2}}{1 + m _{1} m _{2}}

5. Shortest Distance Between Two Curves

The shortest distance between two non-intersecting differentiable curves is measured along their common normal: the segment joining the nearest points is perpendicular to both curves (wherever the normals are defined).

Curve and a line. The nearest point $P (x_{1}, y_{1})$ of the curve is where the tangent is parallel to the line. Solve $f^{'} (x_{1}) =$ slope of the line, then take the perpendicular distance from $P$ to the line. First check that they do not intersect (otherwise the distance is $0$ ).

Figure 13: The closest point

P

of the parabola to the line is where the tangent is parallel to the line:

2 x + 3 = 1

, so

P (- 1, 0)

P F

is then a common normal and the shortest distance is

\frac{∣ - 1 - 0 - 2∣}{2} = \frac{3}{2}

Point and a circle. For a point $A$ outside a circle with centre $C$ and radius $r$ , the nearest and farthest points of the circle lie on line $A C$ : distances $∣ A C ∣ - r$ and $∣ A C ∣ + r$ . For two circles that do not meet, the shortest distance is $∣ C_{1} C_{2} ∣ - r_{1} - r_{2}$ along the line of centres.

Figure 14: Mind map of the geometrical applications of derivatives.

6. Solved Examples

Solved Example 1

Find the equation of the tangent to

y = e^{x}

x = 0

. Hence draw the graph.

Solution:

At $x = 0$ : $y = e^{0} = 1$ , so the point of contact is $(0, 1)$ .
$\frac{d y}{d x} = e^{x}$ , so the slope at $x = 0$ is $1$ .
Tangent: $y - 1 = 1 (x - 0)$ .

Answer: $y = x + 1$ (Figure 1: the curve lies above this tangent).

Solved Example 2

Find the equations of all lines that are tangent to

y = \frac{1}{x - 1}

and parallel to the line

x + y = 0

Solution:

Required slope $= - 1$ . Let the contact point be $(x_{1}, y_{1})$ : $\frac{d y}{d x} = - \frac{1}{( x _{1} - 1 ) ^{2}} = - 1$ .
$(x_{1} - 1)^{2} = 1$ , so $x_{1} = 0$ or $2$ , giving $y_{1} = - 1$ or $1$ .
Tangents: $y + 1 = - (x - 0)$ and $y - 1 = - (x - 2)$ .

Answer: $x + y + 1 = 0$ and $x + y = 3$ (Figure 3).

Solved Example 3

Find the equation of the normal to the curve

y = ∣ x^{2} - ∣ x ∣∣

x = - 2

Solution:

Near $x = - 2$ : $∣ x ∣ = - x$ and $x^{2} + x = 2 > 0$ , so $y = x^{2} + x$ . Point of contact: $(- 2, 2)$ .
$\frac{d y}{d x} = 2 x + 1 = - 3$ at $x = - 2$ , so the normal has slope $\frac{1}{3}$ .
Normal: $y - 2 = \frac{1}{3} (x + 2)$ .

Answer: $3 y = x + 8$ (Figure 4).

Solved Example 4

Prove that the sum of the intercepts of the tangent at any point of

x + y = a

on the coordinate axes is constant.

Solution:

Differentiate: $\frac{1}{2 x} + \frac{1}{2 y} \frac{d y}{d x} = 0$ , so $\frac{d y}{d x} = - \frac{y}{x}$ .
Tangent at $P (x_{1}, y_{1})$ : $y - y_{1} = - \frac{y _{1}}{x _{1}} (x - x_{1})$ . Divide by $y_{1}$ and rearrange: $\frac{x}{x _{1}} + \frac{y}{y _{1}} = x_{1} + y_{1} = a$ .
Intercepts: $A = a x_{1} = a x_{1}$ on the $x$ -axis and $B = a y_{1}$ on the $y$ -axis.
Sum $= a (x_{1} + y_{1}) = a \cdot a$ .

Answer: the sum of intercepts is $a$ , a constant (Figure 10).

Solved Example 5

Find the equations of all possible normals to the parabola

x^{2} = 4 y

drawn from the point

(1, 2)

Solution:

Let the normal meet the parabola at $Q (h, \frac{h ^{2}}{4})$ . Slope of the normal at $Q$ : $- \frac{d x}{d y}_{x = h} = - \frac{2}{h}$ .
The normal passes through $P (1, 2)$ , so slope $P Q = \frac{h ^{2} /4 - 2}{h - 1} = - \frac{2}{h}$ .
$\frac{h ^{3}}{4} - 2 h = - 2 h + 2$ , so $h^{3} = 8$ and the only real root is $h = 2$ .
$Q = (2, 1)$ and the normal is $y - 1 = - 1 (x - 2)$ .

Answer: only one normal, $x + y = 3$ (Figure 7). The cubic $h^{3} = 8$ has a single real root, so only one real normal passes through $(1, 2)$ .

Solved Example 6

Find the value of

c

for which the line joining

(0, 3)

and

(5, - 2)

is a tangent to

y = \frac{c}{x + 1}

Solution:

The line through $A (0, 3)$ and $B (5, - 2)$ has slope $- 1$ : $x + y = 3$ .
Solve with the curve: $3 - x = \frac{c}{x + 1}$ , so $x^{2} - 2 x + (c - 3) = 0$ .
Tangency means equal roots: $D = 4 - 4 (c - 3) = 0$ , so $c = 4$ .
With $c = 4$ : $x^{2} - 2 x + 1 = 0$ , $x = 1$ , $y = 2$ .

Answer: $c = 4$ ; the point of contact is $(1, 2)$ .

Solved Example 7

The tangent at

P (2, 8)

on the curve

y = x^{3}

meets the curve again at

Q

. Find the coordinates of

Q

Solution:

$\frac{d y}{d x} = 3 x^{2} = 12$ at $x = 2$ . Tangent: $y - 8 = 12 (x - 2)$ , i.e. $y = 12 x - 16$ .
Solving with $y = x^{3}$ : $x^{3} - 12 x + 16 = 0$ . Its roots are the $x$ -coordinates of $P$ and $Q$ .
The line touches at $P$ , so $x = 2$ is a repeated root. Let the third root be $h$ . Sum of roots $= 0$ : $2 + 2 + h = 0$ , so $h = - 4$ .

Answer: $Q (- 4, - 64)$ (Figure 6; indeed $x^{3} - 12 x + 16 = (x - 2)^{2} (x + 4)$ ).

Solved Example 8

Find the length of the tangent for the curve

y = x^{3} + 3 x^{2} + 4 x - 1

x = 0

Solution:

$m = \frac{d y}{d x} = 3 x^{2} + 6 x + 4 = 4$ at $x = 0$ , and $k = y (0) = - 1$ .
Length of tangent $= ∣ k ∣ 1 + \frac{1}{m ^{2}} = 1 \cdot 1 + \frac{1}{16}$ .

Answer: $\frac{17}{4}$ .

Solved Example 9

Prove that for the curve

y = b e^{x / a}

the length of the subtangent at any point is constant.

Solution:

At $(x_{1}, y_{1})$ : $m = \frac{b}{a} e^{x_{1} / a} = \frac{y _{1}}{a}$ .
Subtangent $= \frac{y _{1}}{m} = \frac{y _{1}}{y _{1} / a}$ .

Answer: subtangent $= ∣ a ∣$ , the same at every point.

Solved Example 10

For the curve

y = a ln (x^{2} - a^{2})

, show that the sum of the lengths of the tangent and the subtangent at any point is proportional to the product of the coordinates of the point.

Solution:

At $(x_{1}, y_{1})$ : $m = \frac{2 a x _{1}}{x _{1}^{2} - a ^{2}}$ . On this curve $x_{1}^{2} > a^{2}$ .
$1 + \frac{1}{m ^{2}} = 1 + \frac{( x _{1}^{2} - a ^{2} ) ^{2}}{4 a ^{2} x _{1}^{2}} = \frac{( x _{1}^{2} + a ^{2} ) ^{2}}{4 a ^{2} x _{1}^{2}}$ , so length of tangent $= ∣ y_{1} ∣ \frac{x _{1}^{2} + a ^{2}}{2∣ a x _{1} ∣}$ .
Subtangent $= \frac{y _{1}}{m} = ∣ y_{1} ∣ \frac{x _{1}^{2} - a ^{2}}{2∣ a x _{1} ∣}$ .
Sum $= ∣ y_{1} ∣ \frac{2 x _{1}^{2}}{2∣ a x _{1} ∣} = \frac{x _{1} y _{1}}{a}$ .

Answer: the sum equals $\frac{x _{1} y _{1}}{a}$ , proportional to the product $x_{1} y_{1}$ .

Solved Example 11

Find the angle between the curves

y^{2} = 4 x

and

x^{2} = 4 y

. Are they orthogonal?

Solution:

Solving: $(\frac{x ^{2}}{4})^{2} = 4 x$ gives $x = 0$ or $x = 4$ : the curves meet at $(0, 0)$ and $(4, 4)$ .
Slopes: for $y^{2} = 4 x$ , $\frac{d y}{d x} = \frac{2}{y}$ ; for $x^{2} = 4 y$ , $\frac{d y}{d x} = \frac{x}{2}$ .
At $(0, 0)$ : $\frac{2}{y} \to \infty$ (tangent is the $y$ -axis) and $\frac{x}{2} = 0$ (tangent is the $x$ -axis), so $θ = 9 0^{\circ}$ .
At $(4, 4)$ : $m_{1} = \frac{1}{2}$ , $m_{2} = 2$ ; $tan θ = \frac{2 - 0.5}{1 + 2 ( 0.5 )} = \frac{3}{4}$ .

Answer: $9 0^{\circ}$ at $(0, 0)$ and $tan^{- 1} \frac{3}{4}$ at $(4, 4)$ . Not orthogonal, since the angle at $(4, 4)$ is not $9 0^{\circ}$ (Figure 11).

Solved Example 12

Find the angle between the curves

y^{2} = 4 x

and

y = e^{- x /2}

Solution:

The curves do meet: $e^{- x /2}$ falls from $1$ while $4 x$ rises from $0$ . Let them meet at $(x_{1}, y_{1})$ .
For $y^{2} = 4 x$ : $m_{1} = \frac{2}{y _{1}}$ . For $y = e^{- x /2}$ : $m_{2} = - \frac{1}{2} e^{- x_{1} /2} = - \frac{y _{1}}{2}$ .
$m_{1} m_{2} = \frac{2}{y _{1}} \cdot (- \frac{y _{1}}{2}) = - 1$ .

Answer: $θ = 9 0^{\circ}$ . The intersection point itself was never needed.

Solved Example 13

Find all values of

p

for which the equation

p x^{2} = ln x

has exactly one solution.

Solution:

Case $p \leq 0$ : $g (x) = ln x - p x^{2}$ has $g^{'} (x) = \frac{1}{x} - 2 p x > 0$ on $(0, \infty)$ and runs from $- \infty$ to $\infty$ , so there is exactly one root.
Case $p > 0$ : exactly one root only when $y = p x^{2}$ and $y = ln x$ touch. Equal slopes: $2 p x_{1} = \frac{1}{x _{1}}$ , so $x_{1}^{2} = \frac{1}{2 p}$ .
Equal heights: $p x_{1}^{2} = ln x_{1}$ gives $\frac{1}{2} = ln x_{1}$ , so $x_{1} = e$ and $p = \frac{1}{2 x _{1}^{2}} = \frac{1}{2 e}$ .

Answer: $p \in (- \infty, 0] \cup {\frac{1}{2 e}}$ (Figure 12).

Solved Example 14

Find the shortest distance between the line

y = x - 2

and the parabola

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

Solution:

They do not meet: $x^{2} + 3 x + 2 = x - 2$ gives $x^{2} + 2 x + 4 = 0$ with $D = - 12 < 0$ .
Nearest point $P (x_{1}, y_{1})$ : tangent parallel to the line, $2 x_{1} + 3 = 1$ , so $x_{1} = - 1$ , $y_{1} = 0$ .
Perpendicular distance from $(- 1, 0)$ to $x - y - 2 = 0$ : $\frac{∣ - 1 - 0 - 2∣}{2}$ .

Answer: $\frac{3}{2}$ (Figure 13).

Solved Example 15

A curve is given by

x = a t^{2}

y = a t^{3}

. A variable pair of perpendicular lines through the origin

O

meets the curve at

P

and

Q

. Show that the locus of the point of intersection of the tangents at

P

and

Q

4 y^{2} = 3 a x - a^{2}

Solution:

Let $P = (a t_{1}^{2}, a t_{1}^{3})$ , $Q = (a t_{2}^{2}, a t_{2}^{3})$ . Slope of $O P$ is $t_{1}$ , of $O Q$ is $t_{2}$ ; perpendicular, so $t_{1} t_{2} = - 1$ .
$\frac{d y}{d x} = \frac{3 a t ^{2}}{2 a t} = \frac{3 t}{2}$ , so the tangent at parameter $t$ is $2 y = 3 t x - a t^{3}$ .
Subtract the two tangents: $3 x (t_{1} - t_{2}) = a (t_{1}^{3} - t_{2}^{3})$ , so $3 x = a (t_{1}^{2} + t_{1} t_{2} + t_{2}^{2}) = a (s^{2} + 1)$ , where $s = t_{1} + t_{2}$ .
Multiply the first tangent by $t_{2}$ , the second by $t_{1}$ , and subtract: $2 y = a t_{1} t_{2} (t_{1} + t_{2}) = - a s$ , so $s = - \frac{2 y}{a}$ .
Eliminate $s$ : $3 x = a (\frac{4 y ^{2}}{a ^{2}} + 1)$ .

Answer: the locus is $4 y^{2} = 3 a x - a^{2}$ .

Solved Example 16

Prove that the segment of the tangent to

y = \frac{a}{2} ln \frac{a + a ^{2} - x ^{2}}{a - a ^{2} - x ^{2}} - a^{2} - x^{2}

between the

y

-axis and the point of contact has constant length.

Solution:

Let $u = a^{2} - x^{2}$ . Then $\frac{d y}{d u} = \frac{a}{2} (\frac{1}{a + u} + \frac{1}{a - u}) - 1 = \frac{a ^{2}}{x ^{2}} - 1 = \frac{u ^{2}}{x ^{2}}$ , and $\frac{d u}{d x} = - \frac{x}{u}$ .
So $m = \frac{d y}{d x} = - \frac{u}{x} = - \frac{a ^{2} - x ^{2}}{x}$ .
From $P (x_{1}, y_{1})$ to the $y$ -axis the tangent covers a horizontal run of $∣ x_{1} ∣$ , so its length is $∣ x_{1} ∣ 1 + m^{2} = ∣ x_{1} ∣ 1 + \frac{a ^{2} - x _{1}^{2}}{x _{1}^{2}} = ∣ x_{1} ∣ \cdot \frac{a}{∣ x _{1} ∣}$ .

Answer: the length is $a$ at every point (this curve is the tractrix).

Solved Example 17

Find the minimum and maximum values of

(x + 2)^{2} + (y - 1)^{2}

(x - 2)^{2} + (y + 1)^{2} \leq 4

Solution:

$(x + 2)^{2} + (y - 1)^{2}$ is the square of the distance of $(x, y)$ from $A (- 2, 1)$ .
$(x, y)$ lies in the disc with centre $C (2, - 1)$ and radius $2$ . $A C = 16 + 4 = 25 > 2$ , so $A$ is outside.
Nearest and farthest points of the disc lie on line $A C$ : distances $25 - 2$ and $25 + 2$ .
Square them: $(25 \mp 2)^{2} = 24 \mp 85$ .

Answer: minimum $24 - 85 \approx 6.11$ , maximum $24 + 85 \approx 41.89$ .

Solved Example 18

Find the equations of the tangent and the normal to the curve

x = a cos^{3} θ

y = a sin^{3} θ

θ = \frac{π}{4}

Solution:

$\frac{d x}{d θ} = - 3 a cos^{2} θ sin θ$ , $\frac{d y}{d θ} = 3 a sin^{2} θ cos θ$ , so $\frac{d y}{d x} = - tan θ = - 1$ at $θ = \frac{π}{4}$ .
Point: $x = y = a (\frac{1}{2})^{3} = \frac{a}{2 2}$ .
Tangent: $y - \frac{a}{2 2} = - (x - \frac{a}{2 2})$ ; normal: slope $1$ through the same point.

Answer: tangent $x + y = \frac{a}{2}$ ; normal $y = x$ .

Solved Example 19

The equation of the normal to

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

x = 0

is
(A)

x + 3 y = 0

(B)

3 x + y = 0

(C)

x - 3 y = 0

(D)

y = 3 x

Solution:

Answer: (A). The point is $(0, 0)$ ; $\frac{d y}{d x} = 4 x + 3 cos x = 3$ at $x = 0$ , so the normal has slope $- \frac{1}{3}$ : $y = - \frac{x}{3}$ , i.e. $x + 3 y = 0$ . (Option (D) is the tangent.)

Solved Example 20

The point on the curve

y^{2} = x

at which the tangent makes an angle of

\frac{π}{4}

with the

x

-axis is
(A)

(\frac{1}{2}, \frac{1}{4})

(B)

(\frac{1}{4}, \frac{1}{2})

(C)

(4, 2)

(D)

(1, 1)

Solution:

Answer: (B). $2 y \frac{d y}{d x} = 1$ , so $\frac{d y}{d x} = \frac{1}{2 y} = tan \frac{π}{4} = 1$ , giving $y = \frac{1}{2}$ and $x = y^{2} = \frac{1}{4}$ . Option (A) is not even on the curve.

Solved Example 21

Find the angle between the curves

y = x^{2}

and

y = (x - 2)^{2}

at their point of intersection.

Solution:

$x^{2} = (x - 2)^{2}$ gives $x = 1$ : the curves meet at $(1, 1)$ .
$m_{1} = 2 x = 2$ ; $m_{2} = 2 (x - 2) = - 2$ .
$tan θ = \frac{2 - ( - 2 )}{1 + ( 2 ) ( - 2 )} = \frac{4}{- 3} = \frac{4}{3}$ .

Answer: $θ = tan^{- 1} \frac{4}{3} \approx 53. 1^{\circ}$ .

Solved Example 22

Find the shortest distance between the parabola

y^{2} = 4 x

and the line

y = x + 3

Solution:

No intersection: $(x + 3)^{2} = 4 x$ gives $x^{2} + 2 x + 9 = 0$ , $D = - 32 < 0$ .
Tangent parallel to the line: $\frac{d y}{d x} = \frac{2}{y} = 1$ , so $y = 2$ , $x = 1$ .
Distance from $(1, 2)$ to $x - y + 3 = 0$ : $\frac{∣1 - 2 + 3∣}{2} = \frac{2}{2}$ .

Answer: $2$ .

Practice Questions

Find the slope of the normal to the curve $x = 1 - a sin θ$ , $y = b cos^{2} θ$ at $θ = \frac{π}{2}$ .Answer: $- \frac{a}{2 b}$ (as the limit, since $\frac{d x}{d θ}$ and $\frac{d y}{d θ}$ both vanish there)
Find the tangent and normal to (i) $y = x^{4} - 6 x^{3} + 13 x^{2} - 10 x + 5$ at $(1, 3)$ (ii) $y^{2} = \frac{x ^{3}}{4 - x}$ at $(2, - 2)$ .Answer: (i) $y = 2 x + 1$ , $x + 2 y = 7$ (ii) $2 x + y = 2$ , $x - 2 y = 6$
Prove that the area of the triangle formed by any tangent to $x y = c^{2}$ and the coordinate axes is constant.Answer: area $= 2 c^{2}$
How many tangents can be drawn from the origin to $y = (x + 1)^{3}$ ? Find them.Answer: two: $y = 0$ and $4 y = 27 x$ (Figure 5)
Find the tangents to $y = \frac{x + 9}{x + 5}$ that pass through the origin.Answer: $x + y = 0$ and $x + 25 y = 0$
For the curve $x^{m + n} = a^{m - n} y^{2 n}$ ( $a > 0$ ; $m, n$ positive integers), prove that the $m$ th power of the subtangent varies as the $n$ th power of the subnormal.Answer: subtangent $= \frac{2 n x}{m + n}$ , subnormal $= \frac{( m + n ) y ^{2}}{2 n x}$ ; both powers are constant multiples of $x^{m}$
Find the length of the subnormal to $y^{2} = x^{3}$ at $(4, 8)$ .Answer: $24$
Find the angle of intersection of (i) $y = x^{2}$ and $6 y = 7 - x^{3}$ at $(1, 1)$ (ii) $x^{2} - y^{2} = 5$ and $\frac{x ^{2}}{18} + \frac{y ^{2}}{8} = 1$ .Answer: $\frac{π}{2}$ in both cases

Common Mistakes to Avoid

Watch out

Using $f^{'} (x)$ instead of its value $f^{'} (x_{1})$ as the slope: the tangent must be a straight line with a number as slope.
Taking the normal's slope as $\frac{1}{m}$ or $- m$ . It is $- \frac{1}{m}$ .
Writing the normal as ' $y - y_{1} = \infty$ ' when $f^{'} (x_{1}) = 0$ . A zero slope gives the normal $x = x_{1}$ ; an infinite slope gives the tangent $x = x_{1}$ .
For a tangent from an outside point $(a, b)$ , using $f^{'} (a)$ : the point is not on the curve. Use the unknown contact point $(h, f (h))$ .
Differentiating $∣ g (x) ∣$ without first fixing the sign of $g$ near the point.
Finding the angle between curves at only one point when they meet at several, or reporting the obtuse angle: use the modulus and check every intersection.
Dropping the modulus in lengths: subtangent $\frac{k}{m}$ and subnormal $∣ k m ∣$ are lengths, never negative.
Finding a 'shortest distance' between curves that actually intersect (answer $0$ ), or forgetting to subtract the radius for a circle.

Frequently Asked Questions

How do you find the equation of the tangent and normal to a curve?

Find the slope $m = f^{'} (x_{1})$ at the point $(x_{1}, y_{1})$ . The tangent is $y - y_{1} = m (x - x_{1})$ and the normal is $y - y_{1} = - \frac{1}{m} (x - x_{1})$ . If $m = 0$ the tangent is horizontal and the normal is the vertical line $x = x_{1}$ .

What is the geometrical meaning of dy/dx?

$\frac{d y}{d x}$ at a point is the slope of the tangent to the curve there, equal to $tan ψ$ where $ψ$ is the angle the tangent makes with the positive $x$ -axis. It is the limit of the slopes of chords through the point.

How do you find the tangent from an external point to a curve?

Take the unknown point of contact $(h, f (h))$ . The tangent there has slope $f^{'} (h)$ and must pass through the given point $(a, b)$ , so solve $f^{'} (h) = \frac{f ( h ) - b}{h - a}$ . Each real root gives one tangent.

What are subtangent and subnormal?

They are the projections on the $x$ -axis of the tangent and normal segments from the point to the $x$ -axis. For a point $(h, k)$ with slope $m$ , the subtangent is $\frac{k}{m}$ and the subnormal is $∣ k m ∣$ .

How do you find the angle between two curves?

Find their point of intersection, compute the slopes $m_{1}$ and $m_{2}$ of the two tangents there, and use $tan θ = \frac{m _{1} - m _{2}}{1 + m _{1} m _{2}}$ . The angle is the acute angle between the tangents.

When are two curves orthogonal?

Two curves are orthogonal when they cut at right angles at every point of intersection, that is when $m_{1} m_{2} = - 1$ at each common point. An ellipse and a hyperbola with the same foci are always orthogonal.

How are tangents and normals asked in JEE Main?

JEE Main asks for tangent or normal equations at a point, tangents parallel or perpendicular to a line, the point where a normal cuts the curve again, angles between curves and lengths of subtangent and subnormal. Most are single-step slope questions mixed with conics.

What tangent and normal problems come in JEE Advanced?

JEE Advanced asks common tangents, tangents from an external point, loci of intersections of tangents, the number of roots of an equation using touching curves, orthogonal families and shortest distances between curves, usually combined with coordinate geometry.

Previous year questions on Geometrical Applications of Derivatives

1 question from past papers, each with a step-by-step solution.

JEE Advanced 2026 Paper 1, Mathematics Section 1 Q2

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Practice questions →Take a mock test

1. Tangent and Normal

1.1 Geometrical meaning of dxdy​

1.2 Equations of tangent and normal

1.3 Parametric and implicit curves

1.4 Tangents with a given slope

2. Tangents and Normals from an External Point

2.1 A line meeting a curve

2.2 Normals through an external point

3. Lengths of Tangent, Normal, Subtangent and Subnormal

3.1 Intercepts of a tangent

4. Angle Between Two Curves

4.1 Touching curves and the number of roots

5. Shortest Distance Between Two Curves

6. Solved Examples

Common Mistakes to Avoid

Frequently Asked Questions

How do you find the equation of the tangent and normal to a curve?

What is the geometrical meaning of dy/dx?

How do you find the tangent from an external point to a curve?

What are subtangent and subnormal?

How do you find the angle between two curves?

When are two curves orthogonal?

How are tangents and normals asked in JEE Main?

What tangent and normal problems come in JEE Advanced?

Previous year questions on Geometrical Applications of Derivatives

Ready to master Application Of Derivatives?

1.1 Geometrical meaning of $\frac{d y}{d x}$