# Angle Between Curves at Intersection

> Angle between curves for JEE Mathematics: tan θ = |(m1 − m2)/(1 + m1m2)|, orthogonality, tangency and the confocal conics right-angle classic.

- Canonical URL: https://prepelephant.com/topics/jee/mathematics/angle-curves-intersection
- Exam / course: JEE · Subject: Mathematics
- Publisher: PrepElephant (https://prepelephant.com) — Prepared and reviewed by the PrepElephant Academic Review Team
- First published: 2026-10-02
- Last updated: 2026-10-02
- How to cite: "Angle Between Curves at Intersection", PrepElephant, https://prepelephant.com/topics/jee/mathematics/angle-curves-intersection

## Direct answer

Curves meeting at a point cut at the angle between their tangents there: for slopes m1 and m2, tan θ = |(m1 − m2)/(1 + m1 m2)|, and the intersection is orthogonal exactly when m1 m2 = −1. For implicit curves F(x, y) = 0, the slope comes from dy/dx = −Fx/Fy evaluated at the common point. Two degenerate readings carry the exam weight: if the denominator vanishes, the tangents are perpendicular (θ = 90°); if m1 = m2, the curves touch — tangency, angle zero. A classical bonus worth one statement: confocal ellipse and hyperbola intersect at right angles.

## What you must remember

- **The formula:** tan θ = |(m1 − m2)/(1 + m1m2)|, the same one from straight lines, applied to the two tangent slopes at the intersection point.
- **Orthogonality:** m1 m2 = −1; in circle language the radius to the point is one of the tangents' normals.
- **Tangency:** m1 = m2 at a common point means the curves touch with zero angle — contact of order at least one.
- **Implicit slope:** for F(x, y) = 0, dy/dx = −(∂F/∂x)/(∂F/∂y) at the point — the circle x² + y² = r² has slope −x/y at (x, y).
- **Parabola slope:** y² = 4ax carries slope 2a/y at (x, y), or 1/t² at parameter t.
- **Find every intersection first:** missing a common point means missing a whole part of the answer — solve the pair completely before computing any angle.
- **Confocal classic:** every ellipse and hyperbola from the same pair of foci cut orthogonally, a result JEE Advanced has quoted directly.

## Two parabolas meeting twice

Find the angles at which y = x² meets y = x³. Equating, x² = x³ gives x²(x − 1) = 0, so the common points are (0, 0) and (1, 1) — two intersections, two behaviours. At (1, 1) the slopes are 2 and 3, so tan θ = |(3 − 2)/(1 + 6)| = 1/7 and θ = tan⁻¹(1/7), a shallow cut. At (0, 0) both slopes equal 0: the tangents coincide and the curves touch, x³ running beneath x² throughout (0, 1) — an intersection without a cut. The lesson generalises: whenever the intersection equation has a repeated root, expect tangency at that point, and check the slope equality before declaring an angle. Now the orthogonal case in one line: the parabola y = x² + 1? A cleaner example is the pair y² = x and x² + y² = 2? They meet at (1, 1) and (1, −1); at (1, 1) the parabola's slope is 1/(2y) = 1/2 while the circle's is −1, giving product −1/2 — not orthogonal, and computing rather than guessing decided it.

## How JEE frames it

JEE Main keeps the computation local: two curves, a named point, the angle from the formula, with the standard distractor being the complement (using cot θ values) or the acute-versus-obtuse twin from dropping the modulus. Implicit differentiation items dominate — the slope of the ellipse x²/25 + y²/9 = 1 at (3, 12/5)? comes from dy/dx = −9x/25y — and the angle question then rides on that slope. JEE Advanced goes structural: show that the curves x³ − 3xy² + 2 = 0? More realistically, prove orthogonality for a family, as with confocal conics, or find a member of a curve family cutting a given curve at a prescribed angle — the orthogonal-trajectories theme in disguise. The two habits that hold marks: first solve for all intersections (the y = x², y = x⁴ pair hides its tangency at the origin inside a repeated root), and second, evaluate slopes at the point, not anywhere else — substituting the point into the derivative is where silent sign errors enter.

## Frequently asked questions

### What is the angle between two curves at an intersection point?

The angle between their tangents at that point: tan θ = |(m1 − m2)/(1 + m1m2)| for the two slopes.

### When do two curves intersect orthogonally?

When the product of their slopes at the common point is −1, each tangent perpendicular to the other.

### How do you get the slope of an implicit curve?

Differentiate F(x, y) = 0 to get dy/dx = −Fx/Fy and evaluate at the point — no need to solve for y.

### What does a repeated root in the intersection equation signal?

Tangency: the curves touch at that point with equal slopes and zero angle of intersection.

### Which famous curve families cut at right angles?

Confocal ellipses and hyperbolas — sharing foci forces orthogonal intersection, the coordinate-geometry classic quoted directly in Advanced papers.