Normals to Conics
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Direct answer
The normal to a conic at a point is the line perpendicular to the tangent there, and each conic has a parametric normal form worth memorising: for the parabola y² = 4ax at the point (am², 2am), the normal is y = −mx + 2am + am³; for the ellipse x²/a² + y²/b² = 1 at (a cos θ, b sin θ), it is ax sec θ − by cosec θ = a² − b²; for the hyperbola x²/a² − y²/b² = 1 at (a sec θ, b tan θ), it is ax cos θ + by cot θ = a² + b². A circle's normal always passes through the centre. The parabola's cubic in m is the deepest of these: up to three normals from a point, their feet conormal, with parameters obeying m₁ + m₂ + m₃ = 0.
What you must remember
- Parabola normal: at (am², 2am) on y² = 4ax, the normal is y = −mx + 2am + am³.
- Ellipse normal: ax sec θ − by cosec θ = a² − b² at (a cos θ, b sin θ); it bisects the angle between the focal radii in the reflected-light property.
- Hyperbola normal: ax cos θ + by cot θ = a² + b² at (a sec θ, b tan θ).
- Circle normal: the line through the centre and the point of contact — no formula needed, a fact that simplifies many composite problems.
- Three normals and conormal points: the normal condition from (h, k) is the cubic am³ + (2a − h)m − k = 0, whose roots are the feet parameters with m₁ + m₂ + m₃ = 0 — the identity behind conormal questions.
- Point-inside test: three real normals exist from points inside the evolute of the parabola, one from points outside it — a fact JEE Advanced states as "the number of normals is".
Normals to the parabola
Find the normal to y² = 4x at (1, 2). Here a = 1 and the point (am², 2am) gives m = 1, so the formula yields y = −x + 3. Re-derive it: differentiating y² = 4x gives 2yy' = 4, so the tangent slope at (1, 2) is 1, the normal slope is −1, and the line is y = −x + 3. The two routes must agree, and the calculus route is the safer three seconds.
Now the cubic's power: from the point (2, 4), the cubic (a = 1) is m³ − 4 = 0, with m = 4^(1/3) alone real — a single normal. From (4, 4): the cubic becomes m³ − 2m − 4 = 0, and testing m = 2 gives 8 − 4 − 4 = 0, so m = 2 is one root; factoring yields (m − 2)(m² + 2m + 2) = 0 with a negative discriminant on the quadratic — again one real normal. The count of real roots is the count of normals; Vieta's m₁ + m₂ + m₃ = 0 holds over all three.
Where normals get nasty
The formula's traps are sign-flavoured: the parabola normal carries −mx (perpendicular to the tangent y = mx − am² + 2am), and the hyperbola carries a² + b² where the ellipse carries a² − b² — the classic substitution error. The second trap is parameter mismatch: the ellipse point (a cos θ, b sin θ) must pair with sec θ and cosec θ in the normal; using tan θ forms from the parabola page produces a confidently wrong line. Main asks for one normal at a stated point — calculus is fastest. Advanced asks for counts, conormal-point identities, and light-ray problems, where the normal bisects incidence and reflection at the ellipse. Every one of these reduces to the cubic — write it first, read the question off its coefficients.
Frequently asked questions
What is the equation of the normal to y² = 4ax at (am², 2am)?
y = −mx + 2am + am³ — perpendicular to the tangent y = mx − am² + 2am at that point.
How many normals can be drawn from a point to a parabola?
Up to three: the condition am³ + (2a − h)m − k = 0 is a cubic in m, so one, two (with a double root), or three real normals depending on the point's position.
What are conormal points?
Points on a conic whose normals pass through one common external point; for the parabola their parameters satisfy m₁ + m₂ + m₃ = 0 by Vieta on the normal cubic.
What is the normal to an ellipse at (a cos θ, b sin θ)?
ax sec θ − by cosec θ = a² − b², the line that bisects the angle between the lines joining the point to the two foci.
Why does a circle's normal need no formula?
Because every normal to a circle is the radius through the point of contact: one point (the centre) determines the line.