Find the locus of the middle points of chords of an ellipse x 2 a 2 + y 2 b 2 = 1 whose distance from the centre is of constant length c .

Consider ellipse x 2 a 2 + y 2 b 2 = 1 Clearly, by the graph, we have to find the locus of midpoint of chord A B . Let P ( h , k ) be the midpoint of the chord A B , T = S 1 , (Chord with midpoint P ) x h a 2 + y k b 2 = h 2 a 2 + k 2 b 2 Now, perpendicular distance of chord from origin is c c = - h 2 a 2 + k 2 b 2 h 2 a 4 + k 2 b 4 ⇒ h 2 a 2 + k 2 b 2 2 = c 2 h 2 a 4 + k 2 b 4 For locus, Replace h , k → ( x , y ) Hence, the locus of point P is x 2 a 2 + y 2 b 2 2 = c 2 x 2 a 4 + y 2 b 4

Find the locus of the middle points of chords of an ellipse x 2 a 2 + y 2 b 2 = 1 whose distance from the centre is of constant length c .

Consider ellipse x 2 a 2 + y 2 b 2 = 1 Clearly, by the graph, we have to find the locus of midpoint of chord A B . Let P ( h , k ) be the midpoint of the chord A B , T = S 1 , (Chord with midpoint P ) x h a 2 + y k b 2 = h 2 a 2 + k 2 b 2 Now, perpendicular distance of chord from origin is c c = - h 2 a 2 + k 2 b 2 h 2 a 4 + k 2 b 4 ⇒ h 2 a 2 + k 2 b 2 2 = c 2 h 2 a 4 + k 2 b 4 For locus, Replace h , k → ( x , y ) Hence, the locus of point P is x 2 a 2 + y 2 b 2 2 = c 2 x 2 a 4 + y 2 b 4

Find the locus of the middle points of chords of an ellipse x 2 a 2 + y 2 b 2 = 1 which subtend a right angle at the centre.

Consider the ellipse, x 2 a 2 + y 2 b 2 = 1 … i Let A B is a chord of ellipse and P h , k is the midpoint. We have to find the locus of the point P . Now, equation of chord A B with respect to point P (mid point) is T = S 1 ⇒ x h a 2 + y k b 2 - 1 = h 2 a 2 + k 2 b 2 - 1 ⇒ x h a 2 + y k b 2 h 2 a 2 + k 2 b 2 = 1 . . . ii Use the concept of homogenization of a curve, x 2 a 2 + y 2 b 2 = x h a 2 + y k b 2 h 2 a 2 + k 2 b 2 2 Since, O A and O B are perpendicular pair of straight lines Therefore, coefficient of x 2 + coefficient of y 2 = 0 1 a 2 h 2 a 2 + k 2 b 2 2 - h 2 a 4 + 1 b 2 h 2 a 2 + k 2 b 2 2 - k 2 b 4 = 0 ⇒ h 2 a 2 + k 2 b 2 2 1 a 2 + 1 b 2 = h 2 a 4 + k 2 b 4 ⇒ b 2 h 2 + a 2 k 2 a 2 b 2 2 b 2 + a 2 a 2 b 2 = b 4 h 2 + a 4 k 2 a 4 b 4 To obtain the locus of middle points of chords, Replace h , k → x , y Hence, locus of point P is a 2 + b 2 b 2 x 2 + a 2 y 2 2 = a 2 b 2 b 4 x 2 + a 4 y 2

Find the locus of the middle points of chords of an ellipse x 2 a 2 + y 2 b 2 = 1 which pass through the given point h , k .

Let the middle point of the chord of the ellipse x 2 a 2 + y 2 b 2 = 1 be x 1 , y 1 . The equation of this chord bisected at the point x 1 , y 1 is given by , T = S 1 x x 1 a 2 + y y 1 b 2 = x 1 2 a 2 + y 1 2 b 2 … 1 As equation 1 passes through h , k , we have h x 1 a 2 + k y 1 b 2 = x 1 2 a 2 + y 1 2 b 2 To find the locus, Replace x 1 , y 1 → x , y is, h x a 2 + k y b 2 = x 2 a 2 + y 2 b 2 , Simplifying this, we can write: b 2 x x - h + a 2 y y - k = 0 .

Find the locus of the middle points of chords of an ellipse x 2 a 2 + y 2 b 2 = 1 whose length is constant = 2 c .

Suppose, the co-ordinates of the middle point of the chord P Q of the ellipse x 2 a 2 + y 2 b 2 = 1 be h , k . We know that the parametric co-ordinates of a point on a line that makes an angle θ with the positive x -axis, and which is at a distance of r units from a fixed point h , k (that lies on the line) is h ± r cos θ , k ± r sin θ . As the length of P Q is 2 c , the co-ordinates of P and Q may be taken as ( h + c cos θ , k + c sin θ ) and ( h - c cos ⁡ θ , k - c sin ⁡ θ ) , respectively. As these points lie on the ellipse, we must have h + c cos ⁡ θ 2 a 2 + k + c sin ⁡ θ 2 b 2 = 1 … 1 and h - c cos ⁡ θ 2 a 2 + k - c sin ⁡ θ 2 b 2 = 1 … 2 Now, adding equations 1 and 2 , 2 a 2 h 2 + c 2 cos 2 ⁡ θ + 2 b 2 k 2 + c 2 sin 2 ⁡ θ = 2 Or b 2 h 2 + c 2 cos 2 ⁡ θ + a 2 k 2 + c 2 sin 2 ⁡ θ = a 2 b 2 Or b 2 h 2 + a 2 k 2 - a 2 b 2 + c 2 a 2 sin 2 ⁡ θ + b 2 cos 2 ⁡ θ = 0 … 3 Again, subtracting equation ( 2 ) from ( 1 ) , 4 c h a 2 cos ⁡ θ + 4 c k b 2 sin ⁡ θ = 0 Or sin θ h b 2 = cos θ - k a 2 = 1 ± h 2 b 4 + k 2 a 4 Hence, sin θ = h b 2 ± h 2 b 4 + k 2 a 4 and cos θ = - k a 2 ± h 2 b 4 + k 2 a 2 Substituting the values of sin ⁡ θ and cos ⁡ θ in equation ( 3 ) , we get b 2 h 2 + a 2 k 2 - a 2 b 2 + c 2 h 2 b 4 a 2 h 2 b 4 + k 2 a 4 + k 2 a 4 b 2 h 2 b 4 + k 2 a 4 = 0 Or b 4 h 2 + k 2 a 4 b 2 h 2 + a 2 k 2 - a 2 b 2 + c 2 a 2 b 2 h 2 b 2 + a 2 k 2 = 0 Generalising, the locus of h , k is b 4 x 2 + a 4 y 2 b 2 x 2 + a 2 y 2 - a 2 b 2 + c 2 a 2 b 2 x 2 b 2 + a 2 y 2 = 0

Find the locus of the middle point of chord of an ellipse x 2 a 2 + y 2 b 2 = 1 , the tangents at the ends of which intersect at right angles.

The equation of the chord whose middle point is ( h , k ) will be T = S 1 x h a 2 + y k b 2 = h 2 a 2 + k 2 b 2 … 1 If α , β be the point of intersection of two mutually perpendicular tangent at the extremities of the chord, hence the point α , β lies on the director circle whose equation is x 2 + y 2 = a 2 + b 2 Therefore, α 2 + β 2 = a 2 + b 2 … 2 The equation to the chord of contact of ( α , β ) with respect to ellipse x 2 a 2 + y 2 b 2 = 1 is T = 0 ⇒ α x a 2 + β y b 2 = 1 … 3 Clearly equations ( 1 ) and ( 3 ) represent the same line, therefore α h = β k = 1 h 2 a 2 + k 2 b 2 . On equating, α = a 2 b 2 h b 2 h 2 + a 2 k 2 and β = a 2 b 2 k b 2 h 2 + a 2 k 2 Now, from equation ( 2 ) , a 4 b 4 h 2 + a 4 b 4 k 2 b 2 h 2 + a 2 k 2 2 = a 2 + b 2 Or a 4 b 4 h 2 + k 2 = a 2 + b 2 b 2 h 2 + a 2 k 2 2 Generalizing, the locus of ( h , k ) is a 4 b 4 x 2 + y 2 = a 2 + b 2 b 2 x 2 + a 2 y 2 2 .

Circles of constant radius c are drawn to pass through the ends of a variable diameter of the ellipse x 2 a 2 + y 2 b 2 = 1 . Prove that the locus of their centres is the curve x 2 + y 2 a 2 x 2 + b 2 y 2 + a 2 b 2 = c 2 a 2 x 2 + b 2 y 2 .

The co-ordinates of the ends of any diameters of the ellipse x 2 a 2 + y 2 b 2 = 1 may be taken as ( a cos ϕ , b sin ϕ ) and ( - a cos ϕ , - b sin ϕ ) . Again taking, ( h , k ) as the co-ordinates of the centre of the circle of given radius c , then equation to the circle will be x - h 2 + y - k 2 = c 2 . Since the circle passes through the end points of the given diameter, their co-ordinates must satisfy it. So, a cos ⁡ ϕ - h 2 + b sin ⁡ ϕ - k 2 = c 2 … 1 And - a cos ⁡ ϕ - h 2 + - b sin ⁡ ϕ - k 2 = c 2 … 2 Adding equation ( 1 ) and ( 2 ) , 2 a 2 cos 2 ⁡ ϕ + h 2 + 2 b 2 sin 2 ⁡ ϕ + k 2 = 2 c 2 … 3 And subtracting equation 1 from 2 , 4 a h cos ⁡ ϕ + 4 b k sin ⁡ ϕ = 0 Or - cos ϕ b k = sin ϕ a h = 1 ± b 2 k 2 + a 2 h 2 … 4 Putting the values of sin ⁡ ϕ and cos ⁡ ϕ from equation ( 4 ) in ( 3 ) , we get a 2 b 2 ( h 2 + k 2 ) b 2 k 2 + a 2 h 2 + h 2 + k 2 = c 2 Or h 2 + k 2 a 2 b 2 + a 2 h 2 + b 2 k 2 = c 2 a 2 h 2 + b 2 k 2 Generalizing, the locus of the centre ( h , k ) is x 2 + y 2 a 2 x 2 + b 2 y 2 + a 2 b 2 = c 2 a 2 x 2 + b 2 y 2 .

Triangles are formed by pairs of tangents drawn from any point on the ellipse a 2 x 2 + b 2 y 2 = a 2 + b 2 2 to the ellipse x 2 a 2 + y 2 b 2 = 1 , and their chord of contact. Prove that the orthocenter of each such triangle lies on the ellipse x 2 a 2 + y 2 b 2 = 1 .

Take the points of contact to be a cos α , b sin α and a cos β , b sin β on the given ellipse x 2 a 2 + y 2 b 2 = 1 . The equations of the tangents at these points can be given as, x a cos α + y b sin α = 1 x a cos β + y b sin β = 1 Solving the two equations, we get the points of intersection as a cos α + β 2 cos α - β 2 , b sin α + β 2 cos α - β 2 . Hence, this point lies on the ellipse given as, a 2 x 2 + y 2 b 2 = a 2 + b 2 2 ∴ a 2 · a 2 cos 2 α + β 2 cos 2 α - β 2 + b 2 · b 2 sin 2 α + β 2 cos 2 α - β 2 = a 2 + b 2 2 a 4 cos 2 α + β 2 + b 4 sin 2 α + β 2 = a 2 + b 2 2 cos 2 α - β 2 … 1 The equation of the perpendicular line from a cos β , b sin β on the tangent x a cos α + y b sin α = 1 can be given as, a sin α x - a cos β - b cos α y - b sin β = 0 … 2 Similarly, the equations of the other perpendicular can be given as, a sin β x - a cos α - b cos β y - b sin α = 0 … 3 Solving equations ( 2 ) and ( 3 ) , we get, x = a 2 a 2 + b 2 cos α + β 2 cos α - β 2 y = b 2 a 2 + b 2 sin α + β 2 cos α - β 2 If the orthocenter of this triangle lies on the ellipse x 2 a 2 + y 2 b 2 = 1 , then we must have, a 4 cos 2 α + β 2 + b 4 sin 2 α + β 2 = a 2 + b 2 2 cos 2 α - β 2 (which is true by equation ( 1 ) ) Therefore, the orthocenter of the triangle lies on the ellipse. Hence, proved.

E M B I B E

The Elements of COORDINATE GEOMETRY Part 1 Cartesian Coordinates>Chapter 10 - The Ellipse>EXAMPLES XXXV>Q 10

MEDIUM

JEE Advanced

IMPORTANT

Earn 100

Find the locus of the middle points of chords of an ellipse $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$ whose distance from the centre is of constant length $c$ .

Important Questions on The Ellipse

MEDIUM

JEE Advanced

IMPORTANT

The Elements of COORDINATE GEOMETRY Part 1 Cartesian Coordinates>Chapter 10 - The Ellipse>EXAMPLES XXXV>Q 11

Find the locus of the middle points of chords of an ellipse

\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1

which subtend a right angle at the centre.

MEDIUM

JEE Advanced

IMPORTANT

The Elements of COORDINATE GEOMETRY Part 1 Cartesian Coordinates>Chapter 10 - The Ellipse>EXAMPLES XXXV>Q 12

Find the locus of the middle points of chords of an ellipse

\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1

which pass through the given point

(h, k) .

HARD

JEE Advanced

IMPORTANT

The Elements of COORDINATE GEOMETRY Part 1 Cartesian Coordinates>Chapter 10 - The Ellipse>EXAMPLES XXXV>Q 13

Find the locus of the middle points of chords of an ellipse

\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1

whose length is constant

(= 2 c) .

HARD

JEE Advanced

IMPORTANT

The Elements of COORDINATE GEOMETRY Part 1 Cartesian Coordinates>Chapter 10 - The Ellipse>EXAMPLES XXXV>Q 15

Find the locus of the middle point of chord of an ellipse

\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1

, the tangents at the ends of which intersect at right angles.

HARD

JEE Advanced

IMPORTANT

The Elements of COORDINATE GEOMETRY Part 1 Cartesian Coordinates>Chapter 10 - The Ellipse>EXAMPLES XXXV>Q 16

Prove that the locus of the intersection of normals at the ends of conjugate diameters of an ellipse

\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1

is the curve

2 {(a^{2} x^{2} + b^{2} y^{2})}^{3} = {(a^{2} - b^{2})}^{2} {(a^{2} x^{2} - b^{2} y^{2})}^{2} .

HARD

JEE Advanced

IMPORTANT

The Elements of COORDINATE GEOMETRY Part 1 Cartesian Coordinates>Chapter 10 - The Ellipse>EXAMPLES XXXV>Q 19

Circles of constant radius

c

are drawn to pass through the ends of a variable diameter of the ellipse

\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1

. Prove that the locus of their centres is the curve

(x^{2} + y^{2}) (a^{2} x^{2} + b^{2} y^{2} + a^{2} b^{2}) = c^{2} (a^{2} x^{2} + b^{2} y^{2})

.

HARD

JEE Advanced

IMPORTANT

The Elements of COORDINATE GEOMETRY Part 1 Cartesian Coordinates>Chapter 10 - The Ellipse>EXAMPLES XXXV>Q 24

Prove that the sum of the angles that the four normals drawn from any point to an ellipse make with the axis is equal to the sum of the angles that the two tangents from the same point make with the axis.

HARD

JEE Advanced

IMPORTANT

The Elements of COORDINATE GEOMETRY Part 1 Cartesian Coordinates>Chapter 10 - The Ellipse>EXAMPLES XXXV>Q 25

Triangles are formed by pairs of tangents drawn from any point on the ellipse