Solved Examples of Ellipse:

Example 1:

Find the points on the ellipse x² + 3y² = 6 where the tangent are equally inclined to the axes. Prove also that the length of the perpendicular from the centre on either of these tangents is 2.

Solution:

The given ellipse is x² + 3y² = 6

Or x²/6 + y²/2 = 1 ...... (1)

If the coordinates of the required point on the ellipse (1) be (√6 cos Φ, √2 sin Φ) then the tangent at the point is x/√6 cos Φ + y/√2 sin Φ = 1 ...... (2)

Slope of (2) = (-cos Φ)/√6 ×√2/(sin Φ ) = (-√2)/√6 cot Φ

As the tangents are equally inclined to the axes so we have

-1/√3 cot Φ = + tan 45^o = + 1

Hence, tan Φ = + 1/√3

The coordinates of the required points are

(±√6 × √3/2, ±√2 × 1/2) and (±√6 × √3/2, ±√2 × 1/2)

= (± (3√2)/2, ±1/√2) and (± (3√2)/2, ± 1/√2)

Again the length of perpendicular from (0, 0) and (2),

= (√6.√2)/√(2 cos²Φ + 6 sin²Φ)

= (2√3)/√((2.3/4) + (6.1/4) )

= (2√3)/√3

= 2.

Example 2:

If P be a point on the ellipse x²/a² + y²/b² = 2/c whose ordinate is √2/c, prove that the angle between the tangent at P and SP is tan^-1 (b²/ac), where S is the focus.

Solution:

The given ellipse x²/a² + y²/b² = 2/c ...... (1)

If (x', √(2/c)) be the coordinates of the given point P on the ellipse (1).

Then the tangent at P will be: (xx')/a² + (yy')/b² = 1

(xx')/(a²(2/c)) + (yy' (√(2/c)))/(b² (2/c)) = 1

The slope of tangent at P is (-b² (2/c) x')/(a (2/c)(√2(2/c))) = m₁ (say)

If S be the focus, then slope of PS =  y'/(x'+ae) =√(2/c)/(x' + a√(2/c)e)= m₂ (say)

If angle between the focal distance SP and tangent at P is θ, then

tan θ = (m₂ - m₁)/(1 + m₂ m₁).

(a² (2/c) + b² x'²+ aeb²x' √(2/c))/((a² x' + a³ e√(2/c) - b²x')√(2/c))

point (x',√(2/c)) lies on ellipse (1), we have

b² x'² + a²(2/c) = (2a² b²)/c

and since a² - b² = a² e² so we have

(-(2a² b²)/c + aeb² x' √(2/c))/((a²e²x' + a²e√(2/c)) √(2/c)) = (√(2/c) ab²(√(2/c)

(a + ex')/(a²e(√(2/c)) a+ex')) √(2/c)) = b²/ae.

Hence, θ = tan^-1 b²/ae.

Hence proved.

Example 3:

If P, Q, R are three points on the ellipse x²/a² + y²/b² = 1 whose eccentric angles are θ, Φ and Ψ then find the area of ΔPQR.

Solution:

The coordinates of the gives points P, Q, R on the ellipse x²/a² + y²/b² = 1, will be (a cos θ, b sin θ), (a cos Φ, b sin Φ) and (a cos Ψ, b sin Ψ).

Area of triangle PQR formed by these points

= 1/2 [x₁y₂ - x₂y₁ + x₂y₃ - x₃y₂ + x₃y₁ - x₁y₃]

= 1/2 [ab cos θ sin Φ - ab sin θ cos Φ + ab cos Φ sin Ψ - ab sin Φ cos Ψ + ab in θ cos Ψ - ab cos θ sin Ψ]

= 1/2 ab [2 sin(Φ-θ)/2 cos (Φ-θ)/2 + 2 sin (Ψ-Φ)/2 cos (Ψ-Φ)/2 + 2 sin (θ-Φ)/2 cos (θ-Φ)/2]

= ab sin (θ-Φ)/2 (cos (θ + Φ + 2Ψ)/2 cos(Φ-θ)/2)

= 2ab sin (θ-Φ)/2 cos (Φ-Ψ)/2 cos (Ψ-θ)/2.

Example 4:

Find the locus of the extremities of the latus recta of all ellipses having a given major axis 6a.

Solution:

Let LSI be the latus rectum, C be the centre of the ellipse and the coordinates of L be (x, y) then x = CS = 3 ae ...... (1)

And y = SL = b²/3a = (9a²(1 - e²))/3a = 3a (1 - e²) ...... (2)

Eliminating the variable 'e' from (1) and (2) we get the locus of L.

Hence putting the value of e from (1) and (2), we get

y = 3a(1-x²/9a²)

⇒ x² = 9a² - 3ay

⇒ x² = 3a(3a - y), which is clearly a parabola. Similarly we can show that the locus of L' is x² = 3ay(y + 3a) which is again a parabola.

Example 5:

Find the equation of the normals at the end of the latus rectum of

the ellipse x²/a² + y²/b² = c² and find the condition when each normal through one end of the minor axis.

Solution:

The ellipse x²/a² + y²/b² = c²

⇒ x²/(a² c² + y²/(b² c²) = 1 ...... (1)

Then the end point of the latus rectum is (ace,(b² c)/a)

The normal at this point will be

(x-ace)/(ace/(a² c²)) = (y-b² c/a)/((b² c/a)/(b² c²))

⇒ ((x-ace))/e ac = (y-(b² c)/a) ac

⇒ (x-ace)/e = y-(b² c)/a

If this normal passes through (0, - bc), then, we have

(-ace)/e = -bc-b²/a c

⇒ a = b + a(1 - e²)

⇒ b - ae² = 0

⇒ b/1-e² ⇒ b²/a² = e⁴

⇒ e⁴ + e² = 1. This is required condition.

Example 6:

The circle x² + y² = 4 is concentric with the ellipse x²/7 + y²/3 = 1; prove that the common tangent is inclined to the major axis at an angle 30^o and find its length.

Solution:

The ellipse x²/7 + y²/3 = 1 ...... (1)

The equation to the circle is

x² + y² = 4 ...... (2)

As the line y = mx + √(a²m² + b²)

i.e. y = mx  + √(7m² + 3) ...... (3)

is always the tangent on the ellipse. If this is also a tangent on the circle (2) then length of perpendicular from the centre (0, 0) on the line (1) must be equal to the radius of circle i.e. 2.

Hence, √(7m²  + 3)/√(1+ m²) = 2

⇒ 7m² + 3 = 2²(1 + m²)

⇒ 7m² - 4m² = 4 - 3

⇒ m² = 1/3

⇒ m = + 1/√3

Hence the common tangent to the two curves is inclined at an angle of tan^-1 (+1/√3) i.e. 30^o to the axis.

Note:

We can also prove the above result by using the fact that the line = mx + √(7m²+3) will be tangent to x² + y² = 4 if discriminant of x² + (m + √(7m² + 3))² = 4 is zero.

Let P and Q be the points of contact of the common tangent with ellipse and circle respectively and O be the common centre of the two, then

PQ = √(OP² - OQ²) [∠CPQ = 90^o]

The coordinates of P are [(-a²m)/√(a²m² + b²), b²/√(a²m² + b²)]

i.e. [(-7/√3)/√(16/3), 3/√(16/3)]

i.e.[(-7)/4,(3√3)/4]

and coordinate of O are (0, 0)

So, OP = √(((-7)/4)² + ((3√3)/4)²) = √(19/4)

As OQ = r = 2

.·. PQ = √(OP² - OQ²) = √(19/4 - 4) = √3/2

Example 7:

If q be the angle between CP and normal at point P, on the ellipse a²x² + b²y² = 1, then find out tan θ and prove that its greatest value is (b²-a²)/2ab. C is centre of ellipse and P is any point on ellipse.

Solution:

The equation of the ellipse be

x²/(1/a²) + y²/(1/b²)  = 1 ..... (1)

If θ be the angle between the normal at P = (1/a cos Φ, 1/b sin Φ and PC where C is the centre of the ellipse given by (1) equation to the normal PG is

x/a sec Φ - y/b cosec Φ = 1/a² -1/b²

⇒ bx sec Φ - ay cosec Φ = (b² - a²)/ab

Its slope = (b sec Φ)/(a cosec Φ) = b/a tan Φ = m₁ (say)

The slope of PC = (1/b sin Φ)/(1/a cos Φ) = a/b tan Φ = m₂ (say)

tan θ = (m₁ - m₂)/(1 + m₁m₂)

        = (a/b tan Φ – a/b tan Φ)/(1 + (b/a) tan Φ (a/b) tan Φ)

        = ((b²- a²) tan Φ) /ab(1 – tan² θ)

        = (b² - a²)/2ab. (2tanΦ)/(1 – tan² Φ)

tan θ = (b² - a²)/2ab sin 2Φ

The value of tan θ will be maximum when sin 2Φ is maximum, sin 2Φ is maximum i.e. sin 2Φ is 1. Therefore the greatest value of tan θ is (b²-a²)/2ab.

Example 8:

Find the locus of the point of intersection of the two straight lines (x tan α)/a - y/b + tan x = 0 and x/a + (y tan α)/b where a is fixed angle. Also find the eccentric angle of the point of intersection.

Solution:

Equation of the lines are given as

(x tan α)/a – y/b + tan α = 0 ...... (1)

x/a + (y tan α)/b - 1 = 0 ...... (2)

To find the locus of the point of intersection, we have to eliminate the variable 'tan a' from (1) and (2), so by (2),

y/b = 1/(tan α) (1 – x/a) and by (1)

y/b = tan α (1 + x/a)

Multiplying we get

(y/b)² - (1 – x/a)(1 + x/a)

⇒ x²/a² + y²/b² = 1

This is the equation of an ellipse

Again solving (1) and (2), we get

x = a (1 – tan²α)/((1 + tan²α))

x = a(1 – tan²α)/(sec²α)

Let the abscissa of the point of intersection be a cos φ, then

x = a cos φ = a(1 – tan² α)/(sec² α)

⇒ cos φ = (1 – tan² α)/(sec² α)

⇒ (1 – cos φ )/(1 + cos φ) = (sec²α – (1 – tan²α))/(sec²α + (1 – tan²α)) (By components & dividendo)

                                      = (2 tan² α)/2 = tan²α

⇒ (2 sin²φ/2)/(2 cos² φ/2) = tan² α

⇒ tan² φ/2 = tan² α

⇒ tan φ/2 = tan α

Hence φ = 2α.

Example 9:

If TP and TQ are perpendiculars upon the axes from any point T on the ellipse x²/a² + y²/b² = 4. Prove that PQ is always normal to fixed concentric ellipse.

Solution:

The ellipse is given by x²/(4a²) + y²/(4b²) = 1 ...... (1)

If the co-ordinates of T on the ellipse be (2a cos φ, 2b sin φ) an TP and TQ perpendiculars on x-axis and y-axis respectively, the co-ordinates of P and Q will be (2a cos φ, 0) and (0, 2b sin φ) respectively.

Now equation of PQ is

y - 0 = (2b sin φ – 0) /(0 – 2a cos φ)(x - 2a cos φ)

⇒ x/(2a cos φ) + y/(2b sin φ) = 1

⇒ x/a sec φ + y/b cosec φ = 2 ...... (2)

Now equation to the normal at point (A cos φ, B sin φ) with respect to any other concentric ellipse x²/A²  + y²/B² = 1 is

Ax sec φ - By cosec φ = A² - B² ...... (3)

As (2) and (3) are similar, on comparing then we have,

A/(1⁄2a) + B/(1⁄2b) = A² - B² ...... (4)

Solving the two equations given by (4), we get

B = (2a²b)/(a² - b²) and A = (-2ab²)/(a² - b²)

So the line (2) i.e. x/2a sec φ + y/2 cosec φ = 1 is a normal to the fixed ellipse x²/A² + y²/B².

Where A = (-ab²)/(a²-b²)and B = (a²b)/(a²-b²).

Example 10:

If the straight line y = 2x + 2 meets the ellipse 2x² + 3y² - 6, prove that equation to the circle, described on the line joining the points of intersection as diameters, is 7x² + 7y² + 12x - 4y - 5 = 0.

Soluion:

The line is given as y = 2x + 2 ...... (1)

and the ellipse is given as x²/a² + y²/b² = 1 ...... (2)

Solving (1) and (2), we get

x²/3 + (2x + 2)²/b² = 1

⇒ x²/3 + 2(x² + 2x + 1) = 1

⇒ 7x² + 12x + 6 = 3

⇒ 7x² + 12x + 6 = 3

Let x₁ and x₂ be the two roots of this equation.

x₁ + x₂ = -12/7 ...... (3)

and x₁x₂ = 3/7 ...... (4)

Let y₁ and y₂ be the corresponding ordinates for the abscissa x₁ and x₂; so the co-ordinates of the points of intersection will be (x₁, y₁) and (x₂, y₂).

As these lie on line y = 2(x + 1)

We have y₁ = 2(x₁ + 1) and y₂ = 2(x₂ + 1)

Where y₁ + y₂ = 2(x₁ + x₂) + 4 ...... (5)

y₁y₂ = 4(x₁ + 1) (x₂ + 1) ...... (6)

The equation to the circle drawn with the line joining (x₁, y₁) and (x₂, y₂) as diameter is

(x - x₁) (x - x₂) + (y - y₁) (y - y₂) = 0

⇒x² + y² - x(x₁ + x₂) - y(y₁ + y₂) + x₁x₂ + y₁y₂ = 0

⇒  x² + y² - x(x₁ + x₂) – y [2(x₁ + x₂) + 4] + x₁x₂ + 4[x₁x₂ + (x₁ + x₂) + 1] = 0 ( using (5) and (6))

Putting the values from (3) and (4)

x² + y2 - x (-12/7) - y[2(-12/7) + 4] + 3/7 + 4[3/7 + (-12/7) + 1] = 0

which gives 7x² + 7y² + 12x - 4y - 5 = 0. Hence proved.

Example 11:

If the product of the perpendiculars from the foci upon the polar of P be constant and equal to c². Find the locus of P.

Solution:

Suppose the equation to the ellipse x²/a² + y²/b² = 1. The co-ordinate of foci are (ae, 0) and (-ae, 0).

Let the co-ordinates of P be (h, k). Then polar of P is xh/a² + yk/b² = 1

Or b²xh + a²yk - a²b² = 0 ...... (1)

If P₁ and P₂ be the lengths of the perpendiculars on the line (1) from (ae, 0) respectively are

P₁ = (b²hae – a²b²) /√(b²h² + a⁴k²)

And P₂ = (-a²b² - b²hea)/√(b²h² + a⁴x²)

.·. P₁P₂ = (-b²h²a²e² + a⁴b⁴)/(b⁴h² + a⁴) = c² (By hypothesis)

⇒ a⁴b⁴ - b⁴h⁴a²e² = c²b⁴h² - c²a⁴k²

⇒b⁴h² (c² + a²e²) + c²a⁴k² = a⁴b⁴

Generalizing the locus of the point P(h, x) is

b⁴ x² (c² + a²e²) + c²a⁴y² = a⁴b⁴

Example 12:

Chords of ellipse x²/a² + y²/b² = 1 always touch another concentric ellipse x²/α² + y²/β² = 1, show that the locus of their poles is (α²x²)/a² + (β²y²)/b² = 1.

Solution:

Let (x₁, y₁) be the pole of a chord of the ellipse x²/a² + y²/b² = 1 ...... (1)

Then the equation of this cord is the same as the polar of (x₁, y₁) with respect to (1)

i.e. xx₁/a² + yy₁/b² = 1 ...... (2)

If (2) touches the ellipse x²/α² + y²/β² = 1, then (b²/y₁)² = α²{-{b²x₁/a²y₁}}² + β²

⇒ b⁴/y₁² = (α²b⁴x₁²/a⁴y₁²) + β²

⇒ (α²x₁²)/a⁴ + (β²y₁²)/b⁴ = 1

.·. The locus of (x₁, y₁) is (α²x²)/a² + (β²y²)/b⁴ = 1

Hence proved.

Example 13:

If the straight line y = x tan θ + √((a² tan²θ + b²)/2), θ being the angle of inclination, intersects the ellipse x²/a² + y²/b² = 1, then prove that the straight lines joining the centre to their point of intersection are conjugate diameters.

Solution:

The equation of the ellipse be

x²/a² + y²/b² = 1 ...... (1)

and equation to the line is given as

y = x tan θ + √((a²tan²θ + b²)/2), θ being angle of inclination.

We can write this equation as y = mx + √((a²m² + b²)/2)

⇒ ((y-mx)/√2)/√(a²m² + b² ) = 1 ...... (2)

To get the equation to the lines joining the point of intersection to the origin, making (1) homogeneous with the help of (2), we have

x²/a² + y²/b² = [((y-mx)/√2)/√(a²m² + b²)]² = 2((y² + m²x² - 2mxy))/(a²m² + b²)

⇒ (b²x² + a²y²) (a²m² + b²) = 2a² b²(y² + m²x² - 2mxy) 

y² a²(a²m² - b²) + 4m² - b²xy - b²x² (a²m² - b2) = 0

⇒ y² + (4mb²)/((a²m² + b²)) xy – b²/a²x² = 0 ...... (3)

This equation represents two straight lines y = m₁x and y = m₂x then the combined equation will be y² - (m₁ + m₂)xy + m₁m₂x² = 0.

Comparing (3) and (4); we get

m₁m₂ = -b²/a² which is the condition of diameter to be conjugate. Hence the lines are the conjugate diameters.

Example 14:

The eccentric angles of two points P and Q on the ellipse Φ₁, Φ₂. Find the area of the parallelogram formed by the tangents at the ends of diameters through P and Q.

Solution:

The ellipse is x²/a² + y²/b² = 1

Equation to the tangent at the points P and Q are

x/a cos Φ₁ + y/b sin Φ₁ = 1 ...... (1) and

x/a cos Φ₂ + y/b sin Φ₂ = 1 ...... (2)

Solving (1) and (2), we will have the coordinates of the point of intersection. Multiplying (1) by sin Φ₂ and (2) by sin Φ₁ and subtracting, we get

x/a (sin Φ₂ cos Φ₁ - cos Φ₂ sin Φ₁) = sin Φ₂ - sin Φ₁

⇒ x/a sin (Φ₂ - Φ₁) = 2 sin (Φ₂ - Φ₁)/2 cos (Φ₂ + Φ₁)/2

.·. x = a (cos((Φ₁ - Φ₂)/2))/(cos(Φ₁ - Φ₂)/2) and y = b (sin(Φ₁ + Φ₂)/2)/(cos (Φ₁ - Φ₂)/2)

Above are co-ordinates of the point of intersection L of tangents at p and Q, i.e. at Φ₁ and Φ₂.

Putting Φ₁ = p + Φ₁ in above we get the co-ordinates of the point of intersection M of tangent at Q and P' as

Area of the parallelogram LMNO = 4ΔCLM

= 4.1/2 (x₁y₁ - x₂y₂)

= 2 ab/(sin (Φ₁ - Φ₂)/2 cos (Φ₁ - Φ₂)/2) [- cos2 (Φ₁ + Φ₂)/2 -sin (Φ₁ + Φ₂)/2]

= (-4ab)/(sin(Φ₁ - Φ₂))

= - 4ab cosec (Φ₁ - Φ₂) Area can't be (-) ve

So the area = 4ab cosec (Φ₁ - Φ₂).

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