Ellipse — Tangent and Normal

Stand inside the Gol Gumbaz in Bijapur, Karnataka — one of the largest domes in the world. Whisper something facing the wall on one side, and a person standing at the opposite wall, nearly 40 metres away, hears you clearly. The dome's cross-section is (approximately) an ellipse, and the two whispering positions sit at the two foci. A sound wave leaving one focus strikes the curved wall and bounces to the other focus, every time, at every angle. This happens because of a geometric property of the tangent line: at every point on the ellipse, the tangent makes equal angles with the lines to the two foci. That is the reflection property, and proving it requires the tangent equation.

The tangent and normal to an ellipse come in three forms — point form, parametric form, and slope form — each suited to a different type of problem. All three are derived from one idea: the tangent at a point on a curve is the line with slope equal to the derivative at that point. You already know the ellipse \dfrac{x^2}{a^2} + \dfrac{y^2}{b^2} = 1 and its parametric form (a\cos\theta, b\sin\theta) from the auxiliary circle. Now you will find the tangent and normal in every form that JEE problems demand.

Equation of the tangent: point form

Take the ellipse \dfrac{x^2}{a^2} + \dfrac{y^2}{b^2} = 1 and a point P(x_1, y_1) on it, so \dfrac{x_1^2}{a^2} + \dfrac{y_1^2}{b^2} = 1.

Differentiate the ellipse equation implicitly with respect to x:

Why: implicit differentiation treats y as a function of x. The chain rule gives \frac{d}{dx}(y^2) = 2y \cdot \frac{dy}{dx}, and then you solve for \frac{dy}{dx}.

At the point P(x_1, y_1), the slope of the tangent is m = -\dfrac{b^2 x_1}{a^2 y_1}.

The tangent line through (x_1, y_1) with this slope is:

Multiply both sides by a^2 y_1:

Now divide both sides by a^2 b^2:

The right side equals 1 (because P is on the ellipse). So:

Notice the pattern: in the ellipse equation, replace x^2 with xx_1 and y^2 with yy_1. This "T = S_1" substitution rule (where T is the tangent equation and S_1 is the value of the curve equation at the point) works for all conics — you saw the same pattern for the parabola and the circle.

Equation of the tangent: parametric form

If the point of tangency has eccentric angle \theta, then (x_1, y_1) = (a\cos\theta, b\sin\theta). Substitute directly into the point-form tangent:

This form is clean and symmetric. It uses only \theta, a, and b — no coordinates to remember. When a problem gives you the eccentric angle, this is the form to use.

Condition for tangency (slope form)

Not every problem gives you the point of tangency. Sometimes you know the slope of a line and need to check whether it can be tangent to the ellipse. Take the line y = mx + c. When is it tangent to \dfrac{x^2}{a^2} + \dfrac{y^2}{b^2} = 1?

Substitute y = mx + c into the ellipse equation:

Why: substituting the line into the ellipse eliminates y and gives a quadratic in x. The roots of this quadratic are the x-coordinates of intersection. Tangency means exactly one intersection point — so the discriminant must be zero.

This is a quadratic in x. For the line to be tangent, the discriminant must equal zero:

Divide by 4a^2:

Divide by b^2:

This gives the slope form of the tangent: for any slope m, the tangent lines are

The point of contact can be recovered. Substituting the tangency condition back, the quadratic in x has a double root at x = -\dfrac{a^2 m}{c}, and y = mx + c gives y = \dfrac{b^2}{c}. So the point of contact is \left(-\dfrac{a^2 m}{c}, \;\dfrac{b^2}{c}\right).

Comparison with the parabola. For the parabola y^2 = 4ax, the condition for tangency is c = a/m. For the ellipse, it is c^2 = a^2 m^2 + b^2. The parabola has a single tangent of each slope; the ellipse has two (one with + and one with -), reflecting its closed shape.

Equation of the normal

The normal at any point is perpendicular to the tangent at that point. Since you know the tangent slope, the normal slope is its negative reciprocal.

Normal in point form

At P(x_1, y_1), the tangent has slope m_T = -\dfrac{b^2 x_1}{a^2 y_1}. The normal has slope

The normal line through (x_1, y_1) is:

Multiply through by b^2 x_1:

Since c^2 = a^2 - b^2, divide both sides by x_1 y_1 (assuming neither is zero):

Normal in parametric form

At the point (a\cos\theta, b\sin\theta), substitute x_1 = a\cos\theta and y_1 = b\sin\theta:

Normal in slope form

If the normal has slope m, then m = \dfrac{a^2 y_1}{b^2 x_1}, so y_1 = \dfrac{mb^2 x_1}{a^2}. Substituting into the ellipse equation:

The equation of the normal through (x_1, y_1) with slope m is y - y_1 = m(x - x_1), which gives y = mx - mx_1 + y_1 = mx - mx_1 + \dfrac{mb^2 x_1}{a^2} = mx + x_1 m\left(\dfrac{b^2}{a^2} - 1\right) = mx - \dfrac{m(a^2 - b^2)x_1}{a^2}.

Substituting x_1 = \pm\dfrac{a^2}{\sqrt{a^2 + m^2 b^2}}:

Properties of tangent and normal

The tangent intercepts and tangent length

The tangent at (x_1, y_1) meets the x-axis where y = 0:

It meets the y-axis where x = 0:

So the tangent at any point cuts off a triangle from the coordinate axes with vertices at the origin, \left(\dfrac{a^2}{x_1}, 0\right), and \left(0, \dfrac{b^2}{y_1}\right). The area of this triangle is

This area is not constant — it depends on the point of tangency. But its minimum value occurs at the endpoints of the equal diameters (at \theta = \pi/4), where it equals ab.

The normal passes between the foci

A less obvious property: the normal at any point on the ellipse bisects the angle between the focal radii at that point. That is, if P is a point on the ellipse and F_1, F_2 are the foci, then the normal at P is the angle bisector of \angle F_1PF_2.

This is equivalent to the reflection property: the tangent at P makes equal angles with PF_1 and PF_2. Anything traveling from one focus towards the ellipse and bouncing off (with the angle of incidence equal to the angle of reflection, measured from the tangent) will arrive at the other focus. This is why elliptical rooms work as whispering galleries.

Relation between the normal and the focal radii

Here is a concrete statement of the reflection property. At a point P(x_1, y_1) on the ellipse, the focal radii have lengths:

where e = c/a is the eccentricity. The normal at P divides the segment F_1F_2 internally in the ratio PF_1 : PF_2. By the angle bisector theorem (applied to triangle F_1PF_2), the angle bisector from P to F_1F_2 divides F_1F_2 in the ratio of the adjacent sides. Since the normal bisects the angle at P, the foot of the normal on the major axis is at

You can verify this directly: the normal \dfrac{a^2 x}{x_1} - \dfrac{b^2 y}{y_1} = a^2 - b^2 meets the x-axis (y = 0) at x = \dfrac{(a^2 - b^2)x_1}{a^2} = \dfrac{c^2 x_1}{a^2} = e^2 x_1. So the foot of the normal on the major axis is at x = e^2 x_1, always between the two foci (since e < 1 and |x_1| \leq a, we get |e^2 x_1| < e \cdot a = c).

Worked examples

Common confusions

• "The tangent in point form works for any point." It works only for a point on the ellipse. If (x_1, y_1) is not on the curve, then \frac{xx_1}{a^2} + \frac{yy_1}{b^2} = 1 is not the tangent — it is the equation of the chord of contact (the line joining the two points where tangents from (x_1, y_1) touch the ellipse). That is a different thing, covered in the next article.

• "The normal at the vertex is undefined." At the vertex (a, 0), the point-form normal \frac{a^2 x}{x_1} - \frac{b^2 y}{y_1} = a^2 - b^2 has y_1 = 0 in the denominator. But the normal at the vertex is simply x = a — a vertical line. The parametric form handles this cleanly: at \theta = 0, \sin\theta = 0 and the normal equation \frac{ax}{\cos 0} - \frac{by}{\sin 0} = a^2 - b^2 has a \sin 0 in the denominator, which says the normal is vertical. The formula is not broken; the special case is just a vertical line, which has no finite slope.

• "The condition c^2 = a^2 m^2 + b^2 means c is always large." The intercept c here is the y-intercept of the tangent line, not the focal distance. These are completely different quantities that share a letter. In context, it is always clear which is which: c in the tangency condition is the intercept; c = ae in the ellipse anatomy is the focal distance.

• "For every slope m, there is one tangent." There are two — one with c = +\sqrt{a^2 m^2 + b^2} and one with c = -\sqrt{a^2 m^2 + b^2}. The ellipse is a closed curve, so a line of any slope can touch it on either side.

Going deeper

Proof of the reflection property

The reflection property states: at any point P on the ellipse, the tangent makes equal angles with the focal radii PF_1 and PF_2.

Proof. Take P(a\cos\theta, b\sin\theta) on the ellipse \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 with foci F_1(-c, 0) and F_2(c, 0).

The tangent at P has the equation \frac{x\cos\theta}{a} + \frac{y\sin\theta}{b} = 1, or in standard form: bx\cos\theta + ay\sin\theta = ab.

The distance from F_1(-c, 0) to this line is:

The distance from F_2(c, 0) to this line is:

Now, PF_1 = a + e \cdot a\cos\theta = a + c\cos\theta and PF_2 = a - c\cos\theta (using the focal distance formulas with x_1 = a\cos\theta and e = c/a).

So \dfrac{d_1}{d_2} = \dfrac{a + c\cos\theta}{a - c\cos\theta} = \dfrac{PF_1}{PF_2}.

By the converse of the angle bisector property: a line from a vertex of a triangle that divides the opposite side in the ratio of the adjacent sides is the angle bisector. Here, the tangent (acting as a transversal) creates perpendicular distances from F_1 and F_2 in the ratio PF_1 : PF_2. The angles that PF_1 and PF_2 make with the tangent satisfy \dfrac{\sin\alpha_1}{\sin\alpha_2} = \dfrac{d_1/PF_1}{d_2/PF_2} \cdot \dfrac{PF_2}{PF_1}. Since d_1/PF_1 = d_2/PF_2 (from the ratio above, d_1/d_2 = PF_1/PF_2), we get \sin\alpha_1 = \sin\alpha_2, so \alpha_1 = \alpha_2.

Therefore the tangent makes equal angles with the two focal radii. The normal, being perpendicular to the tangent, bisects the angle \angle F_1PF_2.

Number of tangents from an external point

From a point (h, k) outside the ellipse, you can draw exactly two tangent lines. From a point on the ellipse, the tangent is unique. From a point inside the ellipse, no tangent can be drawn.

The test is the value of S_1 = \dfrac{h^2}{a^2} + \dfrac{k^2}{b^2} - 1:

• S_1 > 0: point is outside — two tangents
• S_1 = 0: point is on the ellipse — one tangent
• S_1 < 0: point is inside — no tangent

This is the same sign test that works for circles (with S = x^2 + y^2 - r^2) and parabolas (with S = y^2 - 4ax). For all conics, the sign of S_1 tells you the position of the point relative to the curve.

The director circle, revisited

From the auxiliary circle article, you know that the director circle is x^2 + y^2 = a^2 + b^2. Here is where it connects to tangents.

Two tangent lines from an external point (h, k) to the ellipse have slopes m_1 and m_2 satisfying the quadratic m^2(h^2 - a^2) - 2mhk + (k^2 - b^2) = 0. By Vieta's formulas, m_1 m_2 = \dfrac{k^2 - b^2}{h^2 - a^2}. The tangents are perpendicular when m_1 m_2 = -1, which gives h^2 + k^2 = a^2 + b^2 — the equation of the director circle.

Where this leads next

You now have the tangent and normal to an ellipse in every form. The next topics build on these equations directly:

• Ellipse — Advanced — chord of contact, chord with a given midpoint, pair of tangents from an external point, and the reflection property applied to real optics.
• Ellipse — Introduction — revisit the standard form, eccentricity, and anatomy if any notation felt unfamiliar.
• Ellipse — Auxiliary Circle and Eccentric Angle — the parametric form and director circle, derived from the auxiliary circle construction.
• Parabola — Tangent and Normal — the same three forms (point, parametric, slope) for the parabola, where the algebra is simpler and the pattern is the same.
• Tangent and Normal to a Circle — the circle is an ellipse with a = b; the tangent formulas reduce to the circle case as a check.