Tangent and Normal

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:

1. 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.

2. 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.

3. Slope Form: The equation of tangent to ellipse x²/a² + y²/b² = 1 in terms of slope is given by y = mx ± √(a²m² + b²).

4. In general, the line x cos c + y sin c = l is a tangent if l² = c² cos²α + b² sin²α.

5. 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)

1. 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.

2. 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.

3. • 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

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 = 

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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