Tangent and Normal

Table of Content

How do you find a tangent line to an Ellipse?
- Equation of the tangent to ellipse in terms of 'm’
- Equation of tangent line to ellipse in different forms
How do you find the equation of the normal to an Ellipse?
Pair of Tangents and Chord of Contact
Related Resources

The line y = mx + c meets the ellipse x²/a² + y²/b² = 1 in two real, coincident or imaginary points according as c² < = or > a²m² + b².

Hence, y = mx + c is tangent to the ellipse x²/a² + y²/b² = 1 if c² = a²m² + b².

The equation of the chord to the ellipse joining two points with eccentric angles α and β is given by

x/a cos ((α + β)/2) + y/b sin ((α + β)/2) = cos ((α - β)/2)

How do you find a tangent line to an Ellipse?

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 (2x₁))/(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 the tangent to 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 tangent line to ellipse in different forms

As illustrated above, the various tangent lines to an ellipse can be summed as:

Point form: As discussed above, the equation of the tangent to the ellipse x²/a² + y²/b² = 1 at the point (x₁, y₁) is xx₁/a² + yy₁/b² = 1.
Parametric form: The equtaion 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.
Slope Form: The equation of tangent to ellipse x²/a² + y²/b² = 1 in terms of slope 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.

How do you find the equation of the normal to an Ellipse?

The normal to a curve is a line perpendicular to the tangent to curve through the point of contact.

As discussed above in case of tanegnts, the various ways of finding the normal are listed below:

A line y = mx + c is normal to the ellipse x²/a² + y²/b² = 1 if c² = m² (a² - b²)²/ (a² + b²m²)

Point form: The equation of the normal to the ellipse x²/a² + y²/b² = 1 at the point (x₁, y₁) is a²x/x₁ - b²y/y₁ = a²- b².
Parametric form: The equtaion of the normal to the ellipse x²/a² + y²/b² = 1 at the point (a cos θ, b sin θ) is ax sec θ - by cosec θ = a² - b².
- ope Form: The equation of normal to ellipse x²/a² + y²/b² = 1 in terms of slope is given by y = mx ± m(a²-b²)/√(a² + b²m²). The coordinates of the point of contact are (((± a²)/√(a²+ b²m²), (± b²m)/√(a²+ b²m²))

How many tangents can be drawn to an ellipse?

We know that y = mx + √(a² m²+b²) is a tangent to the ellipse x²/a² +y²/b² = 1 for real values of m. If this tangent passes through a point (x₁, y₁) we have y₁ = mx₁ + √((a² m²+b²)).

or m²(x₁² - a²) - 2x₁y₁m + (y₁² - b²) = 0, which being a quadratic equation in m gives two values of m. Thus from a point two tangents (corresponding to two values of m) can be drawn to an ellipse.

Pair of Tangents and Chord of Contact

From a fixed point (x₁, y₁) in general two tangents can be drawn to an ellipse. The equation of the pair of tangents drawn to the ellipse x²/a² +y²/b² = 1 is given by (x²/a² + y²/b²-1)((x₁²)/a² +(y₁²)/b²- 1) = (xx₁/a² + yy₁/b²-1)².

In symbols we write SS₁ = T², where

S ≡ x²/a² +y²/b² = 1, S₁≡ (x₁²)/a² +(y₁²)/b² – 1 and T ≡ xx₁/a² + yy₁/b² – 1

Key points on chord of contact:

If from the point P(x₁, y₁) tangents PQ and PR be drawn to the ellipse x²/a²+y²/b² = 1, then the line joining the points of contact Q and R is called the chord of contact. Equation of the chord of contact is xx₁/a² + yy₁/b² - 1 = 0 or T = 0.
Equation of the Chord Joining the Points (α cosα, β sinα), (α cosβ, β sinβ) is x/a cos((α+β)/2)+ y/b sin((α+β)/2) = cos((α-β)/2)
Equation of a chord which is bisected at the point (x₁, y₁) is xx₁/a² + yy₁/b² -1 = (x₁²)/a² +(y₁²)/b² - 1 or T = S₁.

To find the length of the chord intercepted by the ellipse x²/a² + y²/b² = 1 on the straight line y = mx + c.

Points of intersection of the ellipse and the line are given by x²/a² +(mx+c)²/b² = 1

i.e. (a² m² + b²)x² + 2a²cmx + a²(c² - b²) = 0 ...... (1)

Therefore the straight line meets the ellipse in two points (real, coincident or imaginary).

If (x₁, y₁) and (x₂, y₂) be the points of intersection, the length of the chord is

√((x₁-x₂)²+ (y₁-y₂)²) = √(1 + m²) = |x₁ - x₂| ...... (2)

(since y₁ - y₂ = m (x₁ - x₂))

where x₁ and x₂ are the roots of the equation (1), and

x₁ + x₂ = -(2a²cm)/(a² m² + b²), x₁ x₂ (a² (c² - b²))/(a² m² + b²) so that

(x₁ - x₂)² = (x₁ + x₂)² - 4x₁ x₂ = (4a² c² m²)/(a² m² + b²)² -(4a²(c² - b²))/((a²m²+ b²)) = (4a² b² (a² m² + b² - c²))/(a² m² + b²)².

Hence the length of the chord is √(((1+m²)4a² b² (a² m² + b² - c²))/(a² m² + b²)²).

i.e. 2ab/(a² m² + b²)²) √((1+m²))(a²) m²)+b²)-c²))).

=> e cos(α+β)/2 = cos(α-β)/2

=> e [cos α/2 cos β/2 + sin α/2 sin β/2] = cos α/2 cos β/2 + sin α/2 sin β/2

=> e [1 - tan α/2 tan β/2] = 1 + tan α/2 tan β/2 => tan α/2 . tanβ/2 = (e-1)/(e+1).

If the chord passes through (-ae, 0) then tanα/2 . tanβ/2 = (e+1)/(e-1).

Illustration:

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.

Note:

The locus of the point of the intersection of two perpendicular tangents to an ellipse is a circle known as the director circle.

Illustration:

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 mid-point 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.

Illustration:

Prove that the product of the lengths of the perpendiculars drawn from the foci to any tangent to the ellipse x²/16 + y²/9 = 1 is equal to 9.

Solution:

For the given ellipse a = 4, b = 3 and hence 9 = 16 (1 - e²)

⇒ e = √7/4. The foci are thus located at (√7,0) and (-√7,0).

Equation of a tangent to the given ellipse is

y = mx + √(16m²+9) (as a = 4, b = 3).

Lengths p₁ and p₂ of the perpendiculars drawn from the foci are

p₁ = (√(16m²+ 9) +√7m)/(1 + m² ) and p₂ = (√(16m²+ 9) - √7m)/(1+m²)

⇒ p₁p₂ = (16m²+ 9 – 7m²)/(1 + m²) = 9(1 + m²)/(1 + m²) = 9.

Note:

The product of lengths of the perpendiculars drawn from the foci to any tangent to the ellipse x²/a² +y²/b² = 1 is b².

Illustration:

If the normals at the end of a latus rectum of the ellipse x²/a² +y²/b² = 1 passes through the extremity of a minor axis, prove that e⁴+e²-1=0.

Solution:

Equation the normal to te given ellipse at

(ae,b²/a) is (x-ae)/(ae/a²) = (y-b²/a)/(b²/ab²).

Since it passes through (0, - b).

- a² = - ab - b³

⇒ a² = ab + a² (1 - e²)

or b = ae² ⇒ b² = a²e⁴

or a² (1 - e²) = a²e² ans so we get a⁴ + e² - 1 = 0.

Illustration:

The tangent and normal at a point P on an ellipse meet the minor axis at A and B. Prove that AB subtends a right angle at each of foci.

Solution:

The equations of the tangents and normal at a point P(x₁, y₁) on the ellipse x²/a²+ y²/b² = 1 are

xx₁/a² + yy₁/b² = 1 ...... (1)

and (x-x₁)/((x₁/a²)) + (y-y₂)/((y₁/b²)) = 1 ...... (2)

Also the minor axis is y-axis i.e. x = 0

Solving (1) and x = 0, we have A (0, b²/y₁)

Solving (2) and x = 0, we have B (0, y₁ - a²y₁/b²)

Let S(ae, 0) be one of the foci of the ellipse. Then the slope of SA = ((b²/y₁) - 0)/(0-ae)= b²/aey₁ = m₁ (say)

And the slope of SB = ([y₁ (a²/b² ) y₁ ]-0)/(0-ae)

= y₁/ae ((a²-b² ))/b²= (y₁ a² e²)/(aeb²) [·.· b² = a²(1 - e²)]

= aey₁/b² = m₂ (say)

Evidently m₁m₂ = -1 ⇒ SA | SB

i.e. AB subtends a right angle at S(ae, 0). Similarly we can show that AB subtends a right angle at the other focus S'(-ae, 0).

Illustration:

Prove that the locus of the middle points of the portion of tangent to the ellipse x²/a² +y²/b² = (a+b) between the axis is the curve a/x² + b/y² = 4/((a+b)).

Solution:

Any tangent to the ellipse x²/a² +y²/b² = 1 , is

y = mx + √(a² m² + b²)

Therefore the tangent to given ellipse is

y = x + √(a(a+b) m²+b(a+b)) ...... (1)

(1) Meets x-axis at {-√(a(a+b) m²+ b(a+b) )/m,0}

(2) Meets y-axis at {0,√(a(a+b) m²+b(a+b))}

Let (x₁, y₁) be the mid points of the portion of the tangent intercepted between the axes, then

x₁ = 1/2 [{-√(a(a+b) m² + b(a+b))/m} + 0]

y₁ = 1/2 [0+{√(a(a+b) m²+b(a+b))} + 0]

⇒ 4x₁² = (a(a+b) m²+ b(a+b))/m²

and 4y₁² = a(a + )m² + b(a + b)

Eliminating m, we get

a(a+b)/(4x₁²) - b(a+b)/(4y₁²) = am²/(am²+b) + b/(am²+b) = 1

.·. The locus of (x₁, y₁) is a/x² + b/y² = 4/(a+b) = 1

Illustration:

If PS₁Q and PS₂R be two focal chord of the ellipse whose two foci are S₁ and S₂and the eccentric angle of P is 'θ' then show that the equation of chord QR is x/a cos θ + y/b.(1+e²)/(1-e²) sin θ + 1 = 0.

Solution:

Let Q be (a cos α, b sin α) and R be (a cos β, b sin β) then the equation of the chord QR is

x/α cos(α+β)/2 + y/b sin(α+β)/2 = cos(α-β)/2

which on simplifying becomes

x/a (1-tan α/2 tan β/2)+y/b (tan α/2+tan β/2) = 1 + tan α/2 tan β/2 ..... (1)

Also, PQ and PR are focal chord thus

tan θ/2 tan α/2 = (e-1)/(e+1) and tan θ/2 tan β/2 = (e+1)/(e-1)

(From previous illustration)

On substituting the values of tan α/2 and β/2 in (1), we get

equation-of-chord

Illustration:

Show the locus of middle points chord of the ellipse x²/a² +y²/b² = 1 which subtend right angle at the centre is x²/a⁴ + y²/b⁴ =(1/a² +1/b² ) (x²/a² + y²/b²)².

Solution:

Let (x₁, y₁) be the middle point of chord PQ, then its equation is

T = S₁ or ₁/a² + yy₁/b² =(x₁/b²)/a² +(y₁/b²)/b² ...... (1)

Since the origin 'O' is the centre so the equation of pair of lines OP and OQ can be obtained by homogenizing the equation of the ellipse x²/a² +y²/b² = 1, with the help of (1), thus

x²/a² +y²/b² = (1)² = locus-of-middle-points-chord-of-the-ellipse

or ((x₁²)/a² +(y₁²)/b²)² (x²/a² +y²/b² ) = (x² x₁²)/a⁴ +(y² y₁²)/b⁴ + (2xyx₁ y₁)/(a²b²).

It represents a pair of perpendicular lines if

1/a² ((x₁²)/a² +(y₁²)/b² )² - (x₁²)/a⁴ + 1/b² ((x₁²)/a² +(y₁²)/b²)²-(y₁²)/b⁴ = 0.

or, (x₁²)/a⁴ +(y₁²)/b⁴ = (1/a² +1/b²)(x²/a⁴ +y²/b⁴)².

So the locus of (x₁, y₁) is

x²/a⁴ + y²/b⁴ = (1/a² + 1/b² ) (x²/a² +y²/b² )².

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