Theorems on ellipse
Theorem - I:

The equation of the chord joining two points (x1, y1) and (x2, y2) on the ellipse S = 0 is S1 + S2 = S12.

Proof:

Let P(x1, y1) and Q(x2, y2) be two points on the ellipse , then S11 = 0 and S22 = 0.

Consider the first degree equation S1 + S2 = S12 which represents a straight line.

Substituting (x1, y1) it becomes S11 + S12 = S12

⇒ 0 + S12 = S12

∴ (x1, y1) satisfy the equation S1 + S2 = S12.

Similarly (x2, y2) also satisfies the equation S1 + S2 = S12.

∴ Equation of the chord PQ will be S1 + S2 = S12.

Theorem - II:

The condition for a straight line y = mx + c to be a tangent to the ellipse is c2 = a2m2 + b2.

Proof:

Given, y = mx + c ----- (i) and

-----(ii)

Substituting (i) in (ii), we get

(a2m2 + b2) x2 + 2a2 mcx + a2 (c2 – b2) = 0 ...... (iii)

The line will touch the ellipse iff the two points are coincident.

⇔ discriminant of (iii) is zero.

⇔ 4a4 c2m2 – 4(a2m2 + b2) a2 (c2 – b2) = 0

⇔ c2 = a2m2 + b2

⇔ c = ±

Theorem - III:

The equation of the tangent to the ellipse S = 0 at P(x1, y1) is S1 = 0.

Proof:

Let P(x1, y1) and Q(x2, y2) be two points on the S = 0 then S11 = 0 and S22 = 0.

By Theorem - I the equation of the chord PQ is S1 + S2 = S12 ...... (1)

Chord PQ becomes the tangent at P when Q approaches to P that is [Q(x2, y2) approaches to P(x1, y1)]

Therefore the equation of the tangent at P obtained by taking limits as (x2, y2) tends to (x1, y1) on either side of (1)

So the equation of tangent at P given by

i.e., S1 + S1 = S11 [∵ S2 → S1 and S12 → S11 as (x2, y2) → (x1, y1)]

i.e., 2S1 = 0

i.e., S1 = 0
Theorem - IV:

Two tangents can be drawn to an ellipse from an external point.

Proof:

Let the equation of the ellipse be S = = 0.

Let P(x1, y1) be an external point.

Then S11 > 0 ⇒ > 0

Let be a tangent to the ellipse.

If it passes through P, then y1 = mx1 ±

⇒ y1 – mx1 = ±

⇒ (x12 – a2)m2 = 2x1y1m + (y12 – b2) = 0

This is a quadratic equation in m and hence m has two values.

Discriminant of (i) is (2x1y1)2 – 4(x12 – a2)(y12 – b2)

= 4x12y12 – 4(x12y12 – x12b2 – y12 + a2b2)

= 4(b2x12 + a2y12 – a2b2)

= 4a2b2() > 0

∴ The two values of m are real and different.

Hence there exists two tangents from P to the ellipse.
Theorem - V:

The equation of the normal to the ellipse S = 0 at P(x1, y1) is = a2 – b2 where x1 ≠ 0 and y1 ≠ 0.

Proof:

By the Theorem - III, the equation of the tangent to the ellipse S = 0 at P(x1, y1) is

S1 ≡ = 0.

∴ Slope of the tangent at P =

∴ Slope of the normal at P =

Hence the equation of the normal at P(x1, y1) is (y – y1) = (x – x1)

Simplifying this, we get = a2 – b2.
Theorem - VI:

At most four normals can be drawn from a given point to an ellipse.

Proof:

Equation of the normal at the point P(θ) on the ellipse S = 0 is = a2 – b2.

If this passes through the point (x1, y1) then = a2 – b2 ...... (1)

The equation will give different values of θ for which the normal passes through (x1, y1).

Equation (1) can be written as

ax1 – by1 = a2 – b2.

After simplification we get

by1 tan4 + 2 (ax1 + a2e2) tan3 + 2 (ax1 – a2e2) tan – by1 = 0.

This, being a fourth degree equation, gives four values for tan .
If we consider one of the value as α1, tan = α1, θ = 2 Tan– 1 (α1) and the general value of θ = 2nπ + 2 Tan– 1 (α1), which gives the same point on the ellipse as θ

∴ Corresponding to one value of tan , we get on a point on the ellipse, real or imaginary.
Hence there will be at most four normals to the ellipse from a point.