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In the plane, a curve surrounds the two focal points so that the distances to the focal points are constant at all points on the curve. Circles are thought to be special types of ellipses that have their points of focus in the same place. Lines crossing each other at a point on an ellipse are known as tangents.

The formula for a tangent to the ellipse

  • In the plane, a curve surrounds the two focal points, such that the total distances to the focal points remain constant for every point on the curve. 
  • It is said that a circle is a special type of ellipse that has both focal points at the same location. 
  • Tangents to ellipses are lines that cross each other at a point on the ellipse. 

Tangent to an ellipse

  • The line y = mx + c touches the ellipse x2 / a2 + y2 / b2 = 1, so c2 = a2m2 + b2. Lines y = mx ∓ √[a2m2 + b2] represent the tangents to the ellipse.

An ellipse has the following point form

  • At the point (x1, y1), the equation of the tangent to an ellipse is x1 / a2 + yy1 / b2 = 1.

A tangent to an ellipse has the following parametric form

  • The tangent equation at any point (a cosɸ, b sinɸ) is [x / a] cosɸ + [y / b] sinɸ.

A tangent to an ellipse has the following point of contact

  • Line y = mx ∓ √[a2m2 + b2] touches the ellipse x2 / a2 + y2 / b2 = 1 at (∓a2m / √[a2m2 + b2]) , (∓b2 / √[a2m2 + b2]).

Point of intersection of two tangents at points

  • Since we are given the points, we can find the equation of both the tangents using a formula for the equation of tangent in point form.
  • Now, since we know the equation of tangents, we can find their point of intersection.  

An ellipse can be found by finding a tangent line.

The equation of the chord to the ellipse    

The ellipse is x2/a2 +y2/b2 = 1 … (1)

Let P(x1, y1) and Q(x2, y2) be two points on the ellipse. The equation of the line PQ is,

y – y1 = ((y2 – y1)/(x2 – x1)) (x – x1) … (2)

Since points P and Q lie on (1), we get

(y2 – y1)/(x2 – x1) = (-b2 (x2-x1))/(a2 (y2-y1) )

So (2) becomes

y – y1 = (-b2 (x2-x1))/(a2 (y2 – y1)) (x – x1)

As point Q approaches point P along the ellipse, the line PQ tends to the tangent at P. So, by substituting x1 and y1 for x2 and y2 in the above equation, we have the equation of the tangent at P as

y – y1 = (-b2 (2×1))/(a2 (2y1)) (x – x1)

⇒ (xx1)/a2 + yy1/b2 = (x12)/a2 +(y12)/b2 = 1     [as P lies on (1)]

Hence, the equation of the tangent to x2/a2 + y2/b2 

= 1 at P(x1,y1) is xx1/a2 + yy1/b2 = 1. … (3)

Equation of a tangent to the ellipse in terms of ‘m’

Let the line y = mx + c … (4)

Touch the ellipse x2/a2 +y2/b2 = 1 … (5)

Eliminating y between (4) and (5), we get;

x2(b2 + a2m2) + 2a2mcx + a2(c2 – b2) = 0 … (6)

If (4) touches (5) then the roots of (6) must be coincident i.e. D = 0

i.e. (2a2mc)2 = 4(b2 + a2m2) a2(c2 – b2)

Solving this we get c = +√(a2m2+b2)

So the equation of tangent is

y = mx + √(a2m2+b2) for all real m  … (7)

From the equation (6) and (7) we get the point of contact as ((± a2m)/√(a2m2 + b2 ), (± b2)/√(a2m2 + b2))

Equation of a tangent to the ellipse in different forms

As shown above, the various tangent lines of an ellipse can be expressed as follows:

  • In point form, the equation of the tangent to the ellipse x2/a2 + y2/b2 = 1 at the point (x1, y1) is xx1/a2 + yy1/b2 = 1.
  • The equation of the tangent to the ellipse x2/a2 + y2/b2 = 1 at the point (a cos θ, b sin θ) is x cos θ/a + y sin θ/b = 1.
  • In terms of slope, the equation of tangent to ellipse x2/a2 + y2/b2 = 1 is given by y = mx ± √(a2m2 + b2).
  • In general, the line x cos c + y sin c = l is a tangent if l2 = c2 cos2α + b2 sin2α.
  • lx + my + n = 0  is a tangent if n2 = a2l2 + b2m2.

Equation of a tangent to the ellipse problems

1. Given:

Find the locus of the point of intersection of the tangents to the ellipse x2/a2+y2/b2 = 1 (a > b), which meet at right angles.

Solution:

The line y = mx ±√(a2 m2+b2) is a tangent to the given ellipse for all m. Suppose it passes through (h, k).

⇒ k – mh = √(a2m2 + b2 ) ⇒ k2 + m2h2 – 2hkm = a2m2 + b2

⇒ m2 (h2 – a2) – 2hkm + k2 – b2 = 0.

If the tangents are at right angles, then m1m2 = -1.

⇒ (k2-b2)/(h2-a2 ) = – 1 ⇒ h2 + k2 = a2 + b2.

Hence, the locus of the point (h, k) is x2 + y2 = a2 + b2, which is a circle. This circle is called the Director Circle of the ellipse.

2. Given:

Prove that the locus of the mid-points of the intercepts of the tangents to the ellipse x2/a2 + y2/b2 = 1 = 1, intercepted between the axes, is a2/x2 +b2/y2 = 4.

Solution:

The tangent to the ellipse at any point (a cosθ, b sinθ)(x cosθ)/a + (y sinθ)/b = 1.

 Let it meet the axes in P and Q so that P is (a secθ, 0)

 and Q is (0, b cosecθ). If (h, k) is the midpoint of PQ, then h = (a sec θ)/2.

⇒ cosθ = a/2h and k = (b cosecθ)/2 ⇒ sinθ = b/2k.

 Squaring and adding, we get a2/4h2 + b2/(4k2 ) = 1

 Hence, the locus of (h, k) is a2/x2 + b2/y2 = 4.

Conclusion

These are the equations of a tangent to an ellipse. The line y = mx + c meets the ellipse x2/a2 + y2/b2 = 1 at two real, coincident, or imaginary points according to c2 < = or > a2m2 + b2. In other words, y = mx + c is a tangent to the ellipse x2/a2 + y2/b2 = 1, if c2 = a2 + m2 + b2. As a result, an ellipse connecting two points with eccentric angles α and β can be written as follows: x/a cos ((α + β)/2) + y/b sin ((α + β)/2) = cos ((α – β)/2).