MHT-CET Maths · Applications of Derivative

Angle Between Curves, Orthogonality, and Nearest Distance

The angle between two curves is the angle between their tangents at the point where they meet. Find both slopes at the intersection, feed them into the tan formula, and read off the angle — or set the slope product to minus one for a right-angle (orthogonal) intersection.

All Applications of Derivative notes MHT-CET Maths strategy

Why this matters

This is one of MHT-CET Maths' most reliable Applications-of-Derivative pockets: 8 PYQs sit here (1 HARD, 7 MODERATE), and every one reduces to the same two-step drill — get m1 and m2 at the meeting point, then plug into tanθ = |(m1−m2)/(1+m1m2)|. The orthogonality variant (solve a parameter so m1m2 = −1) recurs almost every year with the y²=6x, 9x²+by²=16 family, and the parallel-tangent trick for the shortest line-to-curve distance rides on the exact same slope-matching idea.

Concept 1 of 5: Tangent Slopes of Two Curves at Their Meeting Point

Two curves that cross share exactly one point there, but each keeps its own direction — its own tangent slope. To compare their directions you first need both slopes AT that shared point. Find the intersection, then evaluate dy/dx of each curve there.

Definition

Given two curves meeting at a point $P(x_0, y_0)$ :

Step 1 — the point. Solve the two curve equations simultaneously to find $P$ . Dividing or substituting one equation into the other usually isolates $x_0$ fast.
Step 2 — the two slopes. Differentiate each curve (explicitly or implicitly) and evaluate at $P$ : call them $m_1$ and $m_2$ .

These two numbers $m_1, m_2$ are everything the angle formulas need. For an implicit curve $F(x,y)=0$ , differentiate term by term and solve for $\dfrac{dy}{dx}$ before substituting the coordinates.

Slope at a point on a curve

m = \left.\dfrac{dy}{dx}\right|_{(x_0,\,y_0)}

mtangent slope of one curve at the shared point
(x_0, y_0)the intersection point, found by solving the two curves together

Worked example

Find the slopes of

y = x^2

and

x = y^2

at their common point

(1,1)

Practice this conceptself-check · 4 quick reps

You need slopes at the SHARED point, not at any point

The angle between two curves is defined only where they intersect. Always solve the two equations together for

P

FIRST, then substitute those coordinates into each dy/dx. Evaluating the slopes at convenient but different points gives a meaningless angle.

Differentiate the implicit curve fully

For a curve like

9x^2 + by^2 = 16

, every term differentiates:

18x + 2by\dfrac{dy}{dx} = 0

, giving

\dfrac{dy}{dx} = -\dfrac{9x}{by}

. Forgetting the

\dfrac{dy}{dx}

factor on the

y

-term drops the slope entirely.

Drill 1 more on tangent slopes of two curves at their meeting point

Concept 2 of 5: The Angle Between Two Curves

Once you have the two tangent slopes at the meeting point, the angle between the curves is just the angle between those two lines. The tangent of that angle comes straight from the slope-difference formula — the same one you use for two straight lines.

Definition

If two curves meet with tangent slopes $m_1$ and $m_2$ at the intersection, the acute angle $\theta$ between them satisfies

\tan\theta = \left|\dfrac{m_1 - m_2}{1 + m_1 m_2}\right|.

\theta = \tan^{-1}\left|\dfrac{m_1 - m_2}{1 + m_1 m_2}\right|

. Notes on the formula:

The modulus guarantees the acute angle — report that unless the question asks otherwise.
If $1 + m_1 m_2 = 0$ the tangent is undefined, meaning $\theta = 90^\circ$ (orthogonal — see the next concept).
For exponential curves $y = a^x$ the slope is $a^x \log a$ ; at their common point $(0,1)$ the slopes are simply $\log a$ .

Angle between two curves

\tan\theta = \left|\dfrac{m_1 - m_2}{1 + m_1 m_2}\right|

m_1, m_2the two tangent slopes at the intersection point
$\theta$ the acute angle between the curves

Worked example

Find the acute angle between

y = 10 - x^2

and

y = 2 + x^2

at their intersection.

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

MHT-CET · 2025 · 19 April Shift I · Q140Moderate

Example 2 · Applications of Derivative · Angle Between Curves and Orthogonality

The angle between the curves

xy= 6

and

x^{2}y= 12

Keep the modulus for the ACUTE angle

The formula

\tan\theta = \left|\dfrac{m_1 - m_2}{1 + m_1 m_2}\right|

uses a modulus so

\theta

comes out acute. Dropping it can hand you a negative tangent (an obtuse angle) — check the answer options: MHT-CET usually lists the acute value.

$\dfrac{d}{dx}a^x = a^x\log a$ , not $x\,a^{x-1}$

For

y = 3^x, y = 7^x

the base is constant and the exponent is the variable — the derivative is

a^x\log a

, so the slopes at

(0,1)

are

\log 3

and

\log 7

. Using the power rule here is a common wipe-out.

Drill 2 more on the angle between two curves

Concept 3 of 5: The Angle a Curve Makes With a Coordinate Axis

A coordinate axis is itself a straight line with a known slope: the X-axis has slope 0, the Y-axis is vertical. So 'the angle a curve makes with the axis' is just the angle between the curve's tangent there and that axis — and against the X-axis it collapses to

\tan\theta = |m|

, the slope itself.

Definition

To find the angle a curve makes with an axis at a given point:

Find the curve's tangent slope $m$ at that point (implicit differentiation if needed).
Against the X-axis (slope $0$ ): $\tan\theta = |m|$ , so $\theta = \tan^{-1}|m|$ . A slope of $1$ gives $45^\circ$ ; slope $0$ means the curve is tangent to the axis ( $\theta = 0$ ); an infinite slope means $\theta = 90^\circ$ .
Against the Y-axis: use $\tan\theta = \left|\dfrac{1}{m}\right|$ (the Y-axis is the perpendicular reference).

When a curve passes through the origin, substitute $(0,0)$ into the implicit derivative to read the slope directly.

Angle a curve makes with the X-axis

\tan\theta = |m| \qquad \theta = \tan^{-1}|m|

mtangent slope of the curve at the point on the axis

Worked example

The curve

x^4 - 2xy^2 + y^2 + 3x - 3y = 0

cuts the X-axis at

(0,0)

. Find the angle.

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

MHT-CET · 2023 · 10th May Shift 1 · Q108Moderate

Example 3 · Applications of Derivative · Angle Between Curves and Orthogonality

The curve

x^4-2xy^2+y^2+3x-3y=0

cuts the X-axis at

(0,0)

at an angle of

Angle with the X-axis is $\tan^{-1}|m|$ , not the angle formula

You do NOT need the full

\left|\dfrac{m_1-m_2}{1+m_1m_2}\right|

here — the axis has slope

0

, so that formula collapses to

\tan\theta = |m|

. Trying to force the two-slope formula wastes time and invites arithmetic slips.

At the origin, most terms die — keep only the linear ones

Substituting

(0,0)

into an implicit derivative kills every term carrying an

x

y

factor. Only the constant-coefficient linear terms (here

3x

and

-3y

) survive, so the slope reads off in one line:

3 - 3\dfrac{dy}{dx} = 0

Concept 4 of 5: Orthogonal Curves and Solving for a Parameter

Two curves cross 'at right angles' (orthogonally) when their tangents are perpendicular there — so the product of the slopes is

-1

. When one curve carries an unknown constant, this single condition

m_1 m_2 = -1

becomes an equation you solve for that constant.

Definition

Curves intersect orthogonally at $P$ when their tangent slopes satisfy

m_1 \, m_2 = -1.

This is exactly the '

1 + m_1 m_2 = 0

' case of the angle formula (

\theta = 90^\circ

). The standard MHT-CET task: one curve has a free parameter (e.g.

b

9x^2 + by^2 = 16

); impose

m_1 m_2 = -1

at the intersection and use the curve equations to eliminate the coordinates, leaving one equation in the parameter. Because the intersection relation (like

y^2 = 6x

) usually appears in

m_1 m_2

, the coordinates cancel and the parameter drops out cleanly.

Orthogonality condition

m_1 \, m_2 = -1

m_1, m_2the two tangent slopes at the point of intersection

Worked example

y^2 = 4x

and

2x^2 + ky^2 = 6

intersect at right angles, find

k

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

MHT-CET · 2025 · 20 April Shift I · Q111Hard

Example 4 · Applications of Derivative · Angle Between Curves and Orthogonality

If the curves

y^{2}= 6x

and

9x^{2}+by^{2}= 16

intersect each other at right angles, then the value of

b

Orthogonal means slope PRODUCT $= -1$ , not slope sum $= 0$

Perpendicular tangents satisfy

m_1 m_2 = -1

(negative reciprocals). Writing

m_1 + m_2 = 0

m_1 = m_2

is a different condition and gives the wrong parameter value.

Let the intersection relation cancel the coordinates

After imposing

m_1 m_2 = -1

you are left with

x, y

in the equation. Don't panic — substitute the simpler curve relation (

y^2 = 6x

) and the coordinates cancel, leaving a clean equation in the parameter alone.

Drill 1 more on orthogonal curves and solving for a parameter

Concept 5 of 5: Shortest Distance From a Line to a Curve (Parallel-Tangent Trick)

The nearest point of a curve to a straight line is where the curve's tangent runs PARALLEL to that line — a small push either way only increases the gap. So set the curve's slope equal to the line's slope, find that point, then use the point-to-line distance formula.

Definition

To find the shortest distance between a line $ax + by + c = 0$ (slope $m_L = -a/b$ ) and a curve:

Step 1. Set the curve's tangent slope equal to the line's slope: $\dfrac{dy}{dx} = m_L$ . Solve for the nearest point $P$ on the curve.
Step 2. Compute the perpendicular distance from $P(x_0, y_0)$ to the line:

d = \dfrac{|ax_0 + by_0 + c|}{\sqrt{a^2 + b^2}}.

This uses the slope idea from the whole subtopic — the tangent's DIRECTION — plus the standard distance formula.

Point-to-line distance (used at the parallel-tangent point)

d = \dfrac{|a x_0 + b y_0 + c|}{\sqrt{a^2 + b^2}}

(x_0, y_0)the point on the curve where its tangent is parallel to the line
a, b, ccoefficients of the line written as $ax + by + c = 0$

Worked example

Find the shortest distance between the line

y - x = 1

and the curve

x = y^2

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

MHT-CET · 2025 · 20 April Shift I · Q105Moderate

Example 5 · Applications of Derivative · Angle Between Curves and Orthogonality

The shortest distance between the line

y-x= 1

and the curve

x=y^{2}

Nearest point is the PARALLEL-tangent point, not the closest-looking one

The minimum gap happens exactly where the curve's tangent is parallel to the line. Guessing a point or plugging in the vertex usually overshoots — set

\dfrac{dy}{dx}

equal to the line's slope and solve.

Rationalise before matching the options

\dfrac{3/4}{\sqrt2} = \dfrac{3}{4\sqrt2} = \dfrac{3\sqrt2}{8}

. MHT-CET options are usually written with a rational denominator, so rationalise or you may not spot your answer in the list.

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 (5)

Tangent Slopes of Two Curves at Their Meeting Point
Slope at a point on a curve
$m = \left.\dfrac{dy}{dx}\right|_{(x_0,\,y_0)}$
The Angle Between Two Curves
Angle between two curves
$\tan\theta = \left|\dfrac{m_1 - m_2}{1 + m_1 m_2}\right|$
The Angle a Curve Makes With a Coordinate Axis
Angle a curve makes with the X-axis
$\tan\theta = |m| \qquad \theta = \tan^{-1}|m|$
Orthogonal Curves and Solving for a Parameter
Orthogonality condition
$m_1 \, m_2 = -1$
Shortest Distance From a Line to a Curve (Parallel-Tangent Trick)
Point-to-line distance (used at the parallel-tangent point)
$d = \dfrac{|a x_0 + b y_0 + c|}{\sqrt{a^2 + b^2}}$

Watch out for (10)

You need slopes at the SHARED point, not at any point
→ Tangent Slopes of Two Curves at Their Meeting Point
Differentiate the implicit curve fully
→ Tangent Slopes of Two Curves at Their Meeting Point
Keep the modulus for the ACUTE angle
→ The Angle Between Two Curves
$\dfrac{d}{dx}a^x = a^x\log a$ , not $x\,a^{x-1}$
→ The Angle Between Two Curves
Angle with the X-axis is $\tan^{-1}|m|$ , not the angle formula
→ The Angle a Curve Makes With a Coordinate Axis
At the origin, most terms die — keep only the linear ones
→ The Angle a Curve Makes With a Coordinate Axis
Orthogonal means slope PRODUCT $= -1$ , not slope sum $= 0$
→ Orthogonal Curves and Solving for a Parameter
Let the intersection relation cancel the coordinates
→ Orthogonal Curves and Solving for a Parameter
Nearest point is the PARALLEL-tangent point, not the closest-looking one
→ Shortest Distance From a Line to a Curve (Parallel-Tangent Trick)
Rationalise before matching the options
→ Shortest Distance From a Line to a Curve (Parallel-Tangent Trick)

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