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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 45o = + 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 30o 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  + √(7m+ 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. 30o 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 + √(7m+ 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 = 90o]

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)+ ((3√3)/4)2) = √(19/4)

As OQ = r = 2

.·. PQ = √(OP- 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 = (m- 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)

        = (b- 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)- (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 – tanα)/(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/a+ 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)/(a- b2) and A = (-2ab2)/(a- 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 7x+ 7y212x - 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] + x1x+ 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 = (-a2b- b2hea)/√(b2h+ a4x2)

.·. P1P2 = (-b2h2a2e+ a4b4)/(b4h+ 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+ y221, 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 x22 + y22 = 1, then (b2/y1)2 = α2{-{b2x1/a2y1}}2 + β2

⇒ b4/y12 2b4x12/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)/((a2m+ 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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