MHT-CET Maths · Applications of Derivative

Tangents, Normals, and the Slope of a Curve

The derivative read geometrically: the slope of the tangent at a point, the perpendicular normal, the special cases where a tangent is horizontal or vertical, and the recurring MHT-CET puzzles that solve for a point or for a curve's constants from tangency conditions.

All Applications of Derivative notes MHT-CET Maths strategy

Why this matters

This subtopic is the whole chapter's workhorse: 35 PYQs sit here, and it is HARD-heavy — roughly a third are HARD, the rest MODERATE, with only a few EASY. The paper reuses a small set of shapes relentlessly: 'normal parallel to a line ⇒ find the point' (the y = x log x family recurs almost every year), parametric tangent/normal, curve-fitting from touch/gradient conditions, and one-line length/intercept/fixed-point facts. Master the negative-reciprocal normal slope, the dx/dy = 0 test for a vertical tangent, and the parametric dy/dx = (dy/dθ)/(dx/dθ), and most of these become reliable marks.

Concept 1 of 9

Slope of a Curve: Tangent Slope and Normal Slope

Intuition

At a point on a curve the derivative dy/dx IS the slope of the tangent line there. The normal is the line perpendicular to the tangent at that same point, so its slope is the negative reciprocal of the tangent slope. Two numbers unlock everything in this subtopic — the slope, and its negative reciprocal.

Definition

For $y = f(x)$ at the point $(x_1, y_1)$ :

Tangent slope: $m = \left.\dfrac{dy}{dx}\right|_{(x_1, y_1)} = f'(x_1)$ .
Normal slope: $-\dfrac{1}{m}$ — the negative reciprocal (perpendicular lines have slopes multiplying to $-1$ ).
The tangent makes angle $\theta = \tan^{-1} m$ with the positive X-axis.
If a tangent (or normal) is parallel to a given line, it has the same slope as that line; if perpendicular to a line of slope $s$ , its slope is $-1/s$ .

Tangent slope and normal slope

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

mslope of the tangent = value of the derivative at the point
-1/mslope of the normal — negative reciprocal of the tangent slope

Worked example

Find the slope of the tangent and of the normal to

y = x^3 - 3x

x = 2

Slope function: $\dfrac{dy}{dx} = 3x^2 - 3$ .
At $x = 2$ : tangent slope $m = 3(4) - 3 = 9$ .
Normal slope $= -\dfrac{1}{9}$ (negative reciprocal).

Answer:Tangent slope

= 9

; normal slope

= -\dfrac{1}{9}

Practice this conceptself-check · 4 quick reps

Try it yourself

A curve

y = f(x)

has

f'(3) = 2

at the point

(3, 5)

. What angle does the tangent make with the X-axis, and what is the normal slope?

Practice — Level 1 (4 reps)

Quick reps to lock in the method. Try each, then check.

1.
Tangent slope to $y = x^2$ at $x = 3$ .
2.
If the tangent slope is $4$ , what is the normal slope?
3.
Normal is parallel to a line of slope $1$ . What is the tangent slope?
4.
Tangent slope to $y = \log x$ at $x = e$ .

From the bank · past-year question

Example 1Applications of DerivativeEASY

If the normal to the curve

y = f(x)

at the point

(3,4)

makes an angle

\frac{3\pi}{4}

with positive X-axis, then

f'(3)

is equal to

[Q132 · Shift 1 · 2022]

Normal slope is the NEGATIVE reciprocal, not the reciprocal or the negative

If the tangent slope is

m

, the normal slope is

-\dfrac{1}{m}

— both the minus sign AND the reciprocal. Writing

1/m

-m

gives a wrong normal line. When the normal itself is given (e.g. its angle with the X-axis), remember the tangent slope is

-1/(\text{normal slope})

: a normal at

\tfrac{3\pi}{4}

has slope

-1

, so

f' = -1/(-1) = 1

, not

-1

"Parallel to a line" copies the slope; "perpendicular" flips it

A tangent/normal parallel to a line has that line's slope. Perpendicular means negative reciprocal. Read whether it is the TANGENT or the NORMAL that is parallel — that decides which of

m

-1/m

equals the line's slope.

Drill 2 more on slope of a curve: tangent slope and normal slope

Concept 2 of 9

Equations of the Tangent and Normal Lines

Intuition

Once you know a point on the curve and the slope there, both lines follow from point-slope form. The tangent uses the slope m; the normal uses the negative reciprocal -1/m. Everything reduces to 'find the point, find the slope, plug in.'

Definition

At $(x_1, y_1)$ on $y = f(x)$ with tangent slope $m = f'(x_1)$ :

Tangent line: $y - y_1 = m(x - x_1)$ .
Normal line: $y - y_1 = -\dfrac{1}{m}(x - x_1)$ .

For an implicit curve, differentiate implicitly to get $\dfrac{dy}{dx}$ , then evaluate at the point. Always simplify the final line to the option's form (usually $Ax + By + C = 0$ ).

Tangent and normal at a point

\text{tangent: } y - y_1 = m(x - x_1) \qquad \text{normal: } y - y_1 = -\dfrac{1}{m}(x - x_1)

Worked example

Find the equation of the tangent to

y = x^2 + 1

(1, 2)

$\dfrac{dy}{dx} = 2x$ ; at $x = 1$ , $m = 2$ .
Point-slope: $y - 2 = 2(x - 1)$ .
Simplify: $y = 2x$ , i.e. $2x - y = 0$ .

Answer:

2x - y = 0

Practice this conceptself-check · 4 quick reps

Try it yourself

Find the equation of the normal to

y = x^2

(2, 4)

Practice — Level 1 (4 reps)

Quick reps to lock in the method. Try each, then check.

1.
Tangent to $y = x^2$ at $(1,1)$ .
2.
Normal to $y = x^3$ at $(1,1)$ .
3.
Tangent to $y = \log x$ at $(1,0)$ .
4.
Tangent to $xy = 1$ at $(1,1)$ (implicit).

From the bank · past-year question

Example 2Applications of DerivativeMODERATE

The normal to the curve

y(x-2)(x-3) = x+6

at the point, where the curve intersects the Y-axis, passes through the point

[Q148 · 2nd May Shift 1 · 2023]

Use the tangent slope for the tangent, the negative reciprocal for the normal

The single most common slip: writing the normal line with the tangent slope. If the question asks for the NORMAL, substitute

-1/m

into point-slope, not

m

Find the point first, then the slope AT that point

Many stems only describe the point ('where the curve crosses the Y-axis', 'where ordinate = abscissa'). Pin down

(x_1, y_1)

exactly before evaluating the derivative — the slope must be computed at that point, not at a generic

x

Concept 3 of 9

Tangent Parallel to the X-axis or Y-axis

Intuition

A horizontal tangent has zero slope, so dy/dx = 0. A vertical tangent has infinite slope — cleaner to say the RECIPROCAL slope is zero, dx/dy = 0. So for a vertical tangent, differentiate with respect to y and set dx/dy = 0 instead of fighting an infinite dy/dx.

Definition

Tangent parallel to the X-axis (horizontal): $\dfrac{dy}{dx} = 0$ . Solve for the point(s).
Tangent parallel to the Y-axis (vertical): $\dfrac{dy}{dx}$ is undefined; equivalently $\dfrac{dx}{dy} = 0$ . For an implicit curve it is usually easiest to differentiate w.r.t. $y$ and set $\dfrac{dx}{dy} = 0$ .

After finding where the slope condition holds, substitute back into the curve to get the actual point.

Horizontal vs vertical tangent

\text{horizontal: } \dfrac{dy}{dx} = 0 \qquad \text{vertical: } \dfrac{dx}{dy} = 0

Worked example

Find the point on

y = x^2 - 4x + 3

where the tangent is parallel to the X-axis.

$\dfrac{dy}{dx} = 2x - 4$ .
Set $= 0$ : $2x - 4 = 0 \Rightarrow x = 2$ .
Then $y = 4 - 8 + 3 = -1$ .

Answer:

(2, -1)

Practice this conceptself-check · 4 quick reps

Try it yourself

Find the point on

x = y^2 - 2y

where the tangent is parallel to the Y-axis.

Practice — Level 1 (4 reps)

Quick reps to lock in the method. Try each, then check.

1.
Where is the tangent to $y = x^2 - 6x$ horizontal?
2.
Condition for a tangent parallel to the Y-axis.
3.
Abscissa where $y = e^{x} + e^{-x}$ has a horizontal tangent.
4.
Where is the tangent to $y = \sin x$ horizontal on $(0, \pi)$ ?

From the bank · past-year question

Example 3Applications of DerivativeMODERATE

The point on the curve

4y^{2}- 4y+ 2x- 1 = 0

at which the tangent becomes parallel toY-axis is

[Q128 · 23 April Shift I · 2025]

Vertical tangent means dx/dy = 0, not dy/dx = 0

For a tangent parallel to the Y-axis, the slope

\dfrac{dy}{dx}

blows up. Instead of setting the (infinite)

dy/dx = 0

, set the reciprocal

\dfrac{dx}{dy} = 0

. On

4y^2 - 4y + 2x - 1 = 0

, differentiating w.r.t.

y

gives

\dfrac{dx}{dy} = 2 - 4y = 0 \Rightarrow y = \tfrac12

— the clean route.

Don't stop at the slope condition — substitute back for the point

Solving

dy/dx = 0

gives you the x-value (or y-value); the answer is the POINT. Plug back into the curve to get the missing coordinate before matching options.

Drill 2 more on tangent parallel to the x-axis or y-axis

Concept 4 of 9

Tangent or Normal at an Axis-Crossing or Special Point

Intuition

Many stems don't hand you the point — they describe it: 'where the curve crosses the Y-axis' (set x = 0), 'where it crosses the X-axis' (set y = 0), or 'where ordinate = abscissa' (set y = x). Locate the point from that description first, then it's a routine tangent/normal line.

Definition

Translate the description into an equation for the point, then proceed as a standard tangent/normal:

Crosses the Y-axis: put $x = 0$ , solve for $y$ .
Crosses the X-axis: put $y = 0$ , solve for $x$ .
Ordinate = abscissa: put $y = x$ into the curve.

Then compute the slope at that point and write the line $y - y_1 = m(x - x_1)$ (tangent) or with slope $-1/m$ (normal).

Locate the special point, then the line

\text{Y-axis: } x = 0 \qquad \text{X-axis: } y = 0 \qquad \text{ordinate = abscissa: } y = x

Worked example

Find the tangent to

y = e^{2x}

at the point where it crosses the Y-axis.

Crosses the Y-axis at $x = 0$ : $y = e^0 = 1$ , so the point is $(0, 1)$ .
$\dfrac{dy}{dx} = 2e^{2x}$ ; at $x = 0$ , $m = 2$ .
Tangent: $y - 1 = 2(x - 0) \Rightarrow 2x - y + 1 = 0$ .

Answer:

2x - y + 1 = 0

Practice this conceptself-check · 4 quick reps

Try it yourself

Find the tangent to

y = 1 - e^{x/3}

at its intersection with the Y-axis.

Practice — Level 1 (4 reps)

Quick reps to lock in the method. Try each, then check.

1.
Point where $y = 2 - x$ with $(1+x^2)y = 2 - x$ crosses the X-axis.
2.
Y-intercept point of $y = be^{-x/a}$ .
3.
On $y = \sqrt{9 - 2x^2}$ , the point with ordinate = abscissa.
4.
Tangent slope of $y = be^{-x/a}$ at $(0, b)$ .

From the bank · past-year question

Example 4Applications of DerivativeMODERATE

The equation of the tangent to the curve

y=be^{-x/a}

at the point where it crosses the Y axis is

[Q150 · 26 April Shift II · 2025]

Read the axis correctly: Y-axis ⇒ x = 0, X-axis ⇒ y = 0

'Crosses the Y-axis' means the x-coordinate is

0

(set

x = 0

); 'crosses the X-axis' means

y = 0

. Swapping these puts you at the wrong point and every later step is wrong.

'Ordinate = abscissa' means y = x, not a numerical guess

Substitute

y = x

into the curve and solve — e.g. on

y = \sqrt{9 - 2x^2}

x = \sqrt{9 - 2x^2} \Rightarrow x = \sqrt{3}

. Then differentiate at that point.

Drill 3 more on tangent or normal at an axis-crossing or special point

Concept 5 of 9

Tangents and Normals to Parametric Curves

Intuition

When x and y are each given in terms of a parameter t (or θ), you get the slope by dividing the two parameter-derivatives: dy/dx = (dy/dt)/(dx/dt). Evaluate that at the given parameter for the slope, get the point by plugging the parameter into x(t) and y(t), then write the line. The second derivative needs an extra chain-rule step.

Definition

For $x = x(t)$ , $y = y(t)$ :

First derivative (slope): $\dfrac{dy}{dx} = \dfrac{dy/dt}{dx/dt}$ .
Second derivative: $\dfrac{d^2y}{dx^2} = \dfrac{\frac{d}{dt}\!\left(\frac{dy}{dx}\right)}{dx/dt}$ — differentiate the slope w.r.t. $t$ , then divide by $dx/dt$ again. Do not differentiate $dy/dx$ w.r.t. $t$ and stop.

Get the point by substituting the parameter into $x(t), y(t)$ , then form the tangent/normal line.

Parametric slope and second derivative

\dfrac{dy}{dx} = \dfrac{dy/dt}{dx/dt} \qquad \dfrac{d^2y}{dx^2} = \dfrac{1}{dx/dt}\cdot\dfrac{d}{dt}\!\left(\dfrac{dy}{dx}\right)

Worked example

Find the equation of the tangent to

x = a\cos^3\theta,\ y = a\sin^3\theta

\theta = \dfrac{\pi}{4}

$\dfrac{dx}{d\theta} = -3a\cos^2\theta\sin\theta$ , $\dfrac{dy}{d\theta} = 3a\sin^2\theta\cos\theta$ .
$\dfrac{dy}{dx} = \dfrac{3a\sin^2\theta\cos\theta}{-3a\cos^2\theta\sin\theta} = -\tan\theta$ ; at $\theta = \tfrac{\pi}{4}$ , slope $= -1$ .
Point: $\left(a\cos^3\tfrac{\pi}{4}, a\sin^3\tfrac{\pi}{4}\right) = \left(\tfrac{a}{2\sqrt2}, \tfrac{a}{2\sqrt2}\right)$ .
Tangent: $y - \tfrac{a}{2\sqrt2} = -1\left(x - \tfrac{a}{2\sqrt2}\right) \Rightarrow x + y = \tfrac{a}{\sqrt2}$ .

Answer:

x + y = \dfrac{a}{\sqrt2}

Practice this conceptself-check · 4 quick reps

Try it yourself

x = 3\tan t

and

y = 3\sec t

, find

\dfrac{d^2y}{dx^2}

t = \dfrac{\pi}{4}

Practice — Level 1 (4 reps)

Quick reps to lock in the method. Try each, then check.

1.
Slope of $x = t^2,\ y = t^3$ at $t = 2$ .
2.
Slope of $x = \cos\theta,\ y = \sin\theta$ at $\theta = \tfrac{\pi}{4}$ .
3.
For $x = t + \sin t,\ y = 1 + \cos t$ , $dy/dx = ?$
4.
First step to get $d^2y/dx^2$ parametrically.

From the bank · past-year question

Example 5Applications of DerivativeHARD

The equation of the tangent to the curve

x = a\cos^3\theta, y = a\sin^3\theta

\theta = \frac{\pi}{4}

[Q130 · 9th May Shift 2 · 2023]

The parametric second derivative has an extra 1/(dx/dt) factor

\dfrac{d^2y}{dx^2} \neq \dfrac{d}{dt}\!\left(\dfrac{dy}{dx}\right)

. You must divide that by

\dfrac{dx}{dt}

again:

\dfrac{d^2y}{dx^2} = \dfrac{1}{dx/dt}\dfrac{d}{dt}\!\left(\dfrac{dy}{dx}\right)

. Forgetting this factor is the classic error on

x = 3\tan t,\ y = 3\sec t

type problems.

Get the point from the parameter, not from x-alone

Substitute the given

t

(or

\theta

) into BOTH

x(t)

and

y(t)

for the point of contact. Then the tangent/normal line uses that point with the parametric slope.

Drill 4 more on tangents and normals to parametric curves

Concept 6 of 9

Normal Parallel (or Perpendicular) to a Given Line: Solve for the Point

Intuition

Here the point of contact is unknown. A parallel/perpendicular condition fixes the slope, and setting the curve's slope equal to that value gives an equation for the point. The signature MHT-CET version is y = x log x: 'normal parallel to a line' fixes the normal slope, hence the tangent slope, hence x.

Definition

When the NORMAL is parallel to a line of slope $s$ : normal slope $= s$ , so the tangent slope $= -\dfrac{1}{s}$ . Set $\dfrac{dy}{dx} = -\dfrac{1}{s}$ and solve for the point, then write the normal line.

If the normal is perpendicular to a line of slope $s$ , the normal slope is $-1/s$ and the tangent slope is $s$ .
The recurring case $y = x\log x$ : $\dfrac{dy}{dx} = 1 + \log x$ . A normal parallel to a slope-1 line needs tangent slope $-1$ , so $1 + \log x = -1 \Rightarrow x = e^{-2}$ , $y = -2e^{-2}$ .

Normal parallel to a line ⇒ tangent slope condition

\text{normal} \parallel \text{line of slope } s \;\Rightarrow\; \dfrac{dy}{dx} = -\dfrac{1}{s}

Worked example

Find the point on

y = x\log x

at which the normal is parallel to the line

2x - 2y = 3

Line slope: $2x - 2y = 3 \Rightarrow y = x - \tfrac32$ , slope $= 1$ . Normal is parallel, so normal slope $= 1$ .
Tangent slope $= -\dfrac{1}{1} = -1$ , i.e. $\dfrac{dy}{dx} = 1 + \log x = -1$ .
$\log x = -2 \Rightarrow x = e^{-2}$ ; $y = e^{-2}\log(e^{-2}) = -2e^{-2}$ .

Answer:

(e^{-2}, -2e^{-2})

Practice this conceptself-check · 4 quick reps

Try it yourself

Find the equation of the normal to

3x^2 - y^2 = 8

parallel to the line

x + 3y = 10

Practice — Level 1 (4 reps)

Quick reps to lock in the method. Try each, then check.

1.
Normal to $y = x\log x$ parallel to $2x - 2y + 3 = 0$ .
2.
If a normal is parallel to a line of slope $1$ , the tangent slope is?
3.
On $y = x\log x$ , the x where the normal has slope $1$ .
4.
Slope of a normal parallel to $x + 3y = 10$ .

From the bank · past-year question

Example 6Applications of DerivativeHARD

The equation of the normal to the curve

y = x\log x

parallel to

2x - 2y + 3 = 0

[Q137 · Shift 1 · 2023]

'Normal parallel to the line' means the NORMAL slope equals the line slope

It's the normal, not the tangent, that matches the given line's slope. So set the tangent slope to the NEGATIVE RECIPROCAL of the line slope. On

y = x\log x

with line slope

1

: tangent slope

= -1

, giving

1 + \log x = -1

. Matching the tangent slope to

1

instead is the standard wrong turn.

For a tangent PARALLEL to a line, match the tangent slope directly

Don't blanket-apply the negative reciprocal. If the TANGENT is parallel to the line, set

\dfrac{dy}{dx} =

line slope. The reciprocal flip is only for a NORMAL-parallel (or tangent-perpendicular) condition. Also watch

y = \cos(x + y)

: differentiate implicitly to

\dfrac{dy}{dx} = \dfrac{-\sin(x+y)}{1 + \sin(x+y)}

before applying the slope condition.

Drill 10 more on normal parallel (or perpendicular) to a given line: solve for the point

Concept 7 of 9

Finding a Curve's Constants from Tangency Conditions

Intuition

Instead of a fixed curve, you're given a curve with unknown constants (a, b, c) plus conditions like 'touches the X-axis at a point' or 'has gradient 3 at the Y-axis'. Each geometric condition becomes one equation; count them until you can solve for the unknowns. 'Touches an axis' is two conditions in one: the point lies on the curve AND the slope there is zero.

Definition

Turn every stated condition into an equation in the unknown constants:

**Passes through $(x_0, y_0)$ :** substitute into the curve.
**Gradient $= m_0$ at a point:** $y'(x_0) = m_0$ .
**Touches the X-axis at $(p, 0)$ :** BOTH $y(p) = 0$ AND $y'(p) = 0$ (a tangent point on the axis is a repeated root — the curve meets and is tangent).

Solve the resulting simultaneous equations for the constants.

Touches the X-axis at (p, 0): two conditions

y(p) = 0 \quad \text{and} \quad y'(p) = 0

Worked example

The curve

y = ax^2 - 6x + b

passes through

(0, 4)

and has a tangent parallel to the X-axis at

x = \tfrac32

. Find

a

and

b

Through $(0, 4)$ : $b = 4$ .
Horizontal tangent at $x = \tfrac32$ : $y' = 2ax - 6 = 0$ at $x = \tfrac32$ gives $3a - 6 = 0 \Rightarrow a = 2$ .

Answer:

a = 2,\ b = 4

Practice this conceptself-check · 4 quick reps

Try it yourself

The curve

y = ax^3 + bx^2 + cx + 2

touches the X-axis at

(1, 0)

and cuts the Y-axis at a point where its gradient is

-3

. Find

a, b, c

Practice — Level 1 (4 reps)

Quick reps to lock in the method. Try each, then check.

1.
'Touches the X-axis at $(p, 0)$ ' gives which two equations?
2.
'Gradient 3 at the Y-axis' for $y = ax^3 + bx^2 + cx + 5$ .
3.
$y = ax^2 - 6x + b$ through $(0,4)$ : value of $b$ .
4.
For the touch-at- $(-2,0)$ cubic, $a + b + c = ?$

From the bank · past-year question

Example 7Applications of DerivativeHARD

The curve

y=ax^3+bx^2+cx+5

touches the x-axis at

(-2,0)

and cuts the y-axis at a point Q where its gradient is 3, then values of

a,b,c

respectively, are

[Q117 · 11th May Shift 2 · 2023]

'Touches the axis' is TWO conditions, not one

A curve that touches (is tangent to) the X-axis at

(p, 0)

satisfies both

y(p) = 0

(point on the axis) and

y'(p) = 0

(slope zero there). Using only

y(p) = 0

loses an equation and you can't solve for all the constants.

'Gradient at the Y-axis' means evaluate y' at x = 0

The gradient where the curve cuts the Y-axis is

y'(0)

. For

y = ax^3 + bx^2 + cx + 5

y'(0) = c

, so 'gradient 3 at the Y-axis' immediately gives

c = 3

— the fastest first equation.

Drill 2 more on finding a curve's constants from tangency conditions

Concept 8 of 9

Tangent Line Given: Solve for the Curve's Parameters

Intuition

The inverse problem: you're told a specific line is a tangent to a curve with unknowns (like y² = px³ + q) at a stated point. Two facts follow — the point lies on the curve, and the curve's slope there equals the given line's slope. Two equations, two unknowns.

Definition

If the line $y = mx + c$ is tangent to a curve with parameters at the point $(x_0, y_0)$ :

Point on the curve: substitute $(x_0, y_0)$ into the curve equation.
Slope match: the curve's derivative at $(x_0, y_0)$ equals $m$ (the line's slope).

Solve the two equations for the parameters. For an implicit curve, differentiate implicitly to get the slope in terms of the parameters.

Given tangent line at a point: two conditions

(x_0, y_0) \text{ on the curve} \qquad \left.\dfrac{dy}{dx}\right|_{(x_0,y_0)} = m_{\text{line}}

Worked example

y = 3x - 1

is a tangent to

y^2 = ax^3 + b

(1, 2)

, find

a

and

b

Slope match: differentiate $y^2 = ax^3 + b$ : $2y\,y' = 3ax^2$ , so $y' = \dfrac{3ax^2}{2y}$ . At $(1, 2)$ : $\dfrac{3a(1)}{4} = 3 \Rightarrow a = 4$ .
Point on curve: $2^2 = a(1^3) + b \Rightarrow 4 = a + b$ . With $a = 4$ : $b = 0$ .

Answer:

a = 4,\ b = 0

Practice this conceptself-check · 4 quick reps

Try it yourself

The slope of the tangent to

xy + ax + by = 0

(1, 1)

is 2. Given the point lies on the curve, find

a - b

Practice — Level 1 (4 reps)

Quick reps to lock in the method. Try each, then check.

1.
Two conditions when a line is tangent to a curve at a point.
2.
For $y^2 = px^3 + q$ at $(2,3)$ with slope 4: value of $p$ .
3.
Same curve/point: value of $q$ .
4.
$p - q$ for that curve.

From the bank · past-year question

Example 8Applications of DerivativeMODERATE

y = 4x-5

is a tangent to the curve

y^2 = px^3+q

(2,3)

, then the values of p and q are respectively

[Q119 · 10th May Shift 2 · 2023]

You need BOTH the point-on-curve equation and the slope equation

One condition alone under-determines the two unknowns. Use

(x_0, y_0)

on the curve for one equation, and the curve's slope at that point equal to the line's slope for the other. Skipping the slope match leaves a free parameter.

Differentiate the curve implicitly, not the line

The slope you match is the CURVE's derivative at the point (in terms of

p, q, a, b

), set equal to the line's known slope. Differentiate

y^2 = px^3 + q

implicitly as

2y\,y' = 3px^2

; don't confuse the line's slope with the curve's derivative expression.

Drill 2 more on tangent line given: solve for the curve's parameters

Concept 9 of 9

Lengths of Tangent/Normal, Intercepts, and Fixed Points

Intuition

A cluster of one-formula results: the length of the tangent/normal segment to the X-axis, the sub-tangent and sub-normal, the sum of the axis intercepts of a tangent, the distance of a normal from the origin, and whether a parametric normal passes through a fixed point. Each is a short plug-in once you know the formula.

Definition

With $y' = \dfrac{dy}{dx}$ at the point of contact:

Length of tangent $= \left|\dfrac{y\sqrt{1 + y'^2}}{y'}\right|$ ; length of normal $= \left|y\sqrt{1 + y'^2}\right|$ .
Sub-tangent $= \left|\dfrac{y}{y'}\right|$ ; sub-normal $= \left|y\,y'\right|$ .
Distance of a line $Ax + By + C = 0$ from the origin $= \dfrac{|C|}{\sqrt{A^2 + B^2}}$ .
Fixed point: if a parametric normal reduces to a form that holds for all $\theta$ , the coordinates independent of $\theta$ give the fixed point.

Length of normal and length of tangent

\ell_{\text{normal}} = \left|y\sqrt{1 + y'^2}\right| \qquad \ell_{\text{tangent}} = \left|\dfrac{y\sqrt{1 + y'^2}}{y'}\right|

yordinate at the point of contact
y'slope at the point of contact

Worked example

Find the sum of the intercepts on the coordinate axes made by the tangent to

\sqrt{x} + \sqrt{y} = \sqrt{a}

Tangent at $(x_1, y_1)$ : $\dfrac{x}{\sqrt{x_1}} + \dfrac{y}{\sqrt{y_1}} = \sqrt{a}$ (from implicit differentiation).
X-intercept $= \sqrt{a\,x_1}$ , Y-intercept $= \sqrt{a\,y_1}$ .
Sum $= \sqrt{a}\left(\sqrt{x_1} + \sqrt{y_1}\right) = \sqrt{a}\cdot\sqrt{a} = a$ , using $\sqrt{x_1} + \sqrt{y_1} = \sqrt{a}$ .

Answer:Sum of intercepts

= a

Practice this conceptself-check · 4 quick reps

Try it yourself

For the curve

x = 2(\cos t + t\sin t),\ y = 2(\sin t - t\cos t)

, find the distance of the normal at parameter

t

from the origin.

Practice — Level 1 (4 reps)

Quick reps to lock in the method. Try each, then check.

1.
Length of the normal in terms of $y$ and $y'$ .
2.
Sub-normal at a point.
3.
Distance of $x + y = 4$ from the origin.
4.
A parametric normal that holds for all $\theta$ passes through a?

From the bank · past-year question

Example 9Applications of DerivativeMODERATE

The sum of intercepts on coordinate axes made by tangent to the curve

\sqrt{x}+\sqrt{y}=\sqrt{a}

[Q122 · 4th May Shift 1 · 2023]

Length of NORMAL and length of TANGENT are different formulas

Length of normal

= \left|y\sqrt{1 + y'^2}\right|

; length of tangent

= \left|\dfrac{y\sqrt{1 + y'^2}}{y'}\right|

— the tangent version has the extra

1/y'

. Using the tangent formula where the normal is asked (or vice versa) is a classic slip; on 'length of normal from a point' problems, pick the one WITHOUT the

1/y'

Distance from the origin uses only the constant term

For a normal written as

Ax + By + C = 0

, the perpendicular distance from the origin is

\dfrac{|C|}{\sqrt{A^2 + B^2}}

. After reducing a parametric normal to

x\cos t + y\sin t = 2

, the distance is

\dfrac{2}{\sqrt{\cos^2 t + \sin^2 t}} = 2

— the

\theta

-dependence cancels.

Drill 4 more on lengths of tangent/normal, intercepts, and fixed points

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (9)

Slope of a Curve: Tangent Slope and Normal Slope
Tangent slope and normal slope
$m_{\text{tangent}} = \left.\dfrac{dy}{dx}\right|_{(x_1,y_1)} \qquad m_{\text{normal}} = -\dfrac{1}{m_{\text{tangent}}}$
Equations of the Tangent and Normal Lines
Tangent and normal at a point
$\text{tangent: } y - y_1 = m(x - x_1) \qquad \text{normal: } y - y_1 = -\dfrac{1}{m}(x - x_1)$
Tangent Parallel to the X-axis or Y-axis
Horizontal vs vertical tangent
$\text{horizontal: } \dfrac{dy}{dx} = 0 \qquad \text{vertical: } \dfrac{dx}{dy} = 0$
Tangent or Normal at an Axis-Crossing or Special Point
Locate the special point, then the line
$\text{Y-axis: } x = 0 \qquad \text{X-axis: } y = 0 \qquad \text{ordinate = abscissa: } y = x$
Tangents and Normals to Parametric Curves
Parametric slope and second derivative
$\dfrac{dy}{dx} = \dfrac{dy/dt}{dx/dt} \qquad \dfrac{d^2y}{dx^2} = \dfrac{1}{dx/dt}\cdot\dfrac{d}{dt}\!\left(\dfrac{dy}{dx}\right)$
Normal Parallel (or Perpendicular) to a Given Line: Solve for the Point
Normal parallel to a line ⇒ tangent slope condition
$\text{normal} \parallel \text{line of slope } s \;\Rightarrow\; \dfrac{dy}{dx} = -\dfrac{1}{s}$
Finding a Curve's Constants from Tangency Conditions
Touches the X-axis at (p, 0): two conditions
$y(p) = 0 \quad \text{and} \quad y'(p) = 0$
Tangent Line Given: Solve for the Curve's Parameters
Given tangent line at a point: two conditions
$(x_0, y_0) \text{ on the curve} \qquad \left.\dfrac{dy}{dx}\right|_{(x_0,y_0)} = m_{\text{line}}$
Lengths of Tangent/Normal, Intercepts, and Fixed Points
Length of normal and length of tangent
$\ell_{\text{normal}} = \left|y\sqrt{1 + y'^2}\right| \qquad \ell_{\text{tangent}} = \left|\dfrac{y\sqrt{1 + y'^2}}{y'}\right|$

Watch out for (18)

Normal slope is the NEGATIVE reciprocal, not the reciprocal or the negative
→ Slope of a Curve: Tangent Slope and Normal Slope
"Parallel to a line" copies the slope; "perpendicular" flips it
→ Slope of a Curve: Tangent Slope and Normal Slope
Use the tangent slope for the tangent, the negative reciprocal for the normal
→ Equations of the Tangent and Normal Lines
Find the point first, then the slope AT that point
→ Equations of the Tangent and Normal Lines
Vertical tangent means dx/dy = 0, not dy/dx = 0
→ Tangent Parallel to the X-axis or Y-axis
Don't stop at the slope condition — substitute back for the point
→ Tangent Parallel to the X-axis or Y-axis
Read the axis correctly: Y-axis ⇒ x = 0, X-axis ⇒ y = 0
→ Tangent or Normal at an Axis-Crossing or Special Point
'Ordinate = abscissa' means y = x, not a numerical guess
→ Tangent or Normal at an Axis-Crossing or Special Point
The parametric second derivative has an extra 1/(dx/dt) factor
→ Tangents and Normals to Parametric Curves
Get the point from the parameter, not from x-alone
→ Tangents and Normals to Parametric Curves
'Normal parallel to the line' means the NORMAL slope equals the line slope
→ Normal Parallel (or Perpendicular) to a Given Line: Solve for the Point
For a tangent PARALLEL to a line, match the tangent slope directly
→ Normal Parallel (or Perpendicular) to a Given Line: Solve for the Point
'Touches the axis' is TWO conditions, not one
→ Finding a Curve's Constants from Tangency Conditions
'Gradient at the Y-axis' means evaluate y' at x = 0
→ Finding a Curve's Constants from Tangency Conditions
You need BOTH the point-on-curve equation and the slope equation
→ Tangent Line Given: Solve for the Curve's Parameters
Differentiate the curve implicitly, not the line
→ Tangent Line Given: Solve for the Curve's Parameters
Length of NORMAL and length of TANGENT are different formulas
→ Lengths of Tangent/Normal, Intercepts, and Fixed Points
Distance from the origin uses only the constant term
→ Lengths of Tangent/Normal, Intercepts, and Fixed Points

Mastery check — 5 interleaved questions

Try each one before clicking. Questions are interleaved across the concepts above, not grouped — interleaving sharpens transfer.

Example 1Applications of DerivativeMODERATE

\alpha x+by+c=0

is normal to curve

xy=1

, then

[Q127 · 10th May Shift 1 · 2024]

Example 2Applications of DerivativeEASY

If the curve

y=ax^{2}- 6x+b

passes through

(0,4)

and has its tangent parallel to the X -axis at

x=\frac{3}{2}

, then the values of

a

and

b

respectively are

[Q133 · 20 April Shift II · 2025]

Example 3Applications of DerivativeMODERATE

The equation of the tangent to the curve

\left( 1 +x^{2} \right)y= 2 -x

, where it crosses the X -axis, is

[Q135 · 26 April Shift I · 2025]

Example 4Applications of DerivativeHARD

If equation of normal to the curve

x = \sqrt{t}

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

t = 4

[Q118 · 11th May Shift 1 · 2023]

Example 5Applications of DerivativeMODERATE

Let C be a curve given by

y(x) = 1 + \sqrt{4x-3}

x > \frac{3}{4}

. If P is a point on C, such that the tangent at P has slope

\frac{2}{3}

, then a point through which the normal at P passes, is

[Q109 · 9th May Shift 1 · 2023]

Drill every past-year question on this subtopic

35 questions from the bank — paginated, with cart and Word-export support.

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Drill every past-year question on this subtopic

Related notes