Diameter of Ellipse

Diameter of Ellipse

Edited By Komal Miglani | Updated on Oct 07, 2024 09:55 AM IST

An ellipse is the set of all points ( x,y ) in a plane such that the sum of their distances from two fixed points is a constant. A line drawn through the center of an ellipse is called the diameter of the ellipse. In real life, we use diamer in orbits, antennas, and artistic compositions.

In this article, we will cover the concept of the Diameter of the Ellipse. This category falls under the broader category of Coordinate Geometry, which is a crucial Chapter in class 11 Mathematics. It is not only essential for board exams but also for competitive exams like the Joint Entrance Examination(JEE Main) and other entrance exams such as SRMJEE, BITSAT, WBJEE, BCECE, and more. A total of seven questions have been asked on JEE MAINS( 2013 to 2023) from this topic.

What is the Diameter of the Ellipse?
The locus of the mid-points of a system of parallel chords of an ellipse is called a diameter and the point where the diameter intersects the ellipse is called the vertex of the diameter.
Locus of Mid Point


Let (h,k) be the mid-point of the chord y=mx+c of the ellipse x2a2+y2b2=1, then
T=S1 [equation of chord bisected at given point]

xha2+ykb2=h2a2+k2 b2k=b2 ha2 m
Hence, the locus of the mid-point is y=b2xa2m


Conjugate Diameters


Two diameters are said to be conjugate when each bisects all chords parallel to the other.

If y=m1x and y=m2x be two conjugate diameters of an ellipse, then

m1m2=b2a2
If PQ and RS are two conjugate diameters. Then the coordinates of the four extremities of two conjugate diameters are

P(acosϕ,bsinϕ)Q(acosϕ,bsinϕ)S(asinϕ,bcosϕ)R(asinϕ,bcosϕ)


Properties of Diameters


1) The tangent at the extremity of any diameter is parallel to the chords it bisects or parallel to the conjugate diameter.
2) The tangents at the ends of any chord meet on the diameter which bisects the chord.

Properties of conjugate diameters
1) The eccentric angles of the ends of a pair of conjugate diameters of an ellipse differ by a right angle.
2) The sum of the square of any two conjugate semi-diameters of an ellipse is constant and equal to the sum of squares of the semi-axis.
3) The product of the focal distances of a point on an ellipse is equal to the square of the semi-diameter which is conjugate to the diameter through the point.
4) Two conjugate diameters are called equi conjugate if their lengths are equal.

Recommended Video Based on Diameter of Ellipse


Solved Examples Based on Diameter of Ellipse


Example 1: The Locus of midpoints of chords of the ellipse x22+y2=1 which are tangents to the ellipse x2+y22=1 is
Solution:
Locus of the mid-point of the chord of the ellipse x22+y2=1is y=x2m
The equation of a tangent to the ellipse x2+y22=1 in slope form is y=mx±m2+2
from eq (i) and eq (ii)

y=(x2y)x+(x2y)2+2(2y2+x2)2=x2+8y2
Hence, the correct answer is (2y2+x2)2=x2+8y2

Example 2: If the product of focal distances of a point P(acosθ,bsinθ) on an ellipse x2a2+y2 b2=1 is λ-times the square of the semi-diameter CD (of conjugate diameter CP ), then λ= Solution: Let foci be S(ae,0) and S(ae,0)AsP is (acosθ,bsinθ)
D is (acos(θ+π2),bsin(θ+π2))

=(asinθ,bcosθ)
By using the definition of an ellipse,

PS=e(PM)
Or PS=e(aeacosθ)

=aaecosθ=a(1ecosθ)( Standard Result ) SP.S'P =a(1ecosθ)a(1+ecosθ)=a2(1e2cos2θ)=a2a2e2cos2θ
But b2=a2(1e2)

SP.S'P =a2(a2b2)cos2θ=a2sin2θ+b2cos2θ=CD2

Hence, the correct answer is 1


Example 3: If one end of the diameter of the ellipse 4x2+y2=16 is (3,2), then the other end is
Solution: Since every diameter of an ellipse passes through the centre and is bisected by it, therefore the coordinates of the other end are (3,2)
Hence, the answer is (3,2)


Example 4: A ray emanating from the point (0,3) is incident on the ellipse 16x2+25y2=400 at the point P with ordinate 4 . Then the equation of the reflected ray after the first reflection is.
Solution: For point P y-coordinate =4
Given ellipse is 16x2+25y2=400

16x2+25(4)2=400,x=0

co-ordinate of co-ordinate of P is (0,4)

e2=11625=925e=35

Foci (±ae,0), i.e. (±3,0)
Equation of reflected ray
(i.e.PS) is x3+y4=1 or 4x+3y=12.


Hence, the correct answer is 4x+3y=12


Example 5: If the points of intersection of the ellipses x2a2+y2b2=1 and x2p2+y2q2=1 be the extremities of the conjugate diameters of first ellipse, then a2P2+b2q2=
Solution: Subtracting in order to find their points of intersection, we get

x2(1a21p2)+y2(1 b21q2)=0
The above equation will represent a pair of conjugate diameters of the first ellipse if

m1 m2=b2a2
But m1 m2=AB=b2a2


(1a21p2)÷(1b21q2)=b2a2 or a2(1a21p2)+b2(1b21q2)=0 or a2p2+b2q2=2
Hence, the correct answer is 2


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