2. Hello!
My Name is Jamil Ahmed
Class-XI A
Roll no-12
Project submitted to:- Suraksha Mam
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3. Ellipse
Definition:- An ellipse is a set of all points in a plane, the
sum of whose distances from two fixed points in the
plane is constant.
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4. 4
✘ Focus:-the two fixed points are called the foci (plural of ‘focus’)
✘ The midpoint of the line segment joining the foci is called the centre of
the ellipse.
✘ The line segment, the foci of the ellipse is called the major axis and the
line segment through the centre and perpendicular to the major axis is
called minor axis.
✘ The end points of the major axis are called the vertices of the ellipse.
5. 5
Relationship between semi-major axis,semi-minor axis and the
distance of the focus from the centre
✘ Length of semi-major axis = a.
✘ Length of semi-minor axis = b.
✘ Distance of the focus from the centre = c
✘ Then,
C=√a²-b² or a²=b²+c²
6. 6
Latus Rectum
✘ Latus rectum of an ellipse is a line segment perpendicular to the major
axis through any of the foci and whose end points lie on the ellipse
✘ Formula of Latus Rectum:-
l=2b²/a
7. 7
Eccentricity (e)
✘ The eccentricity of an ellipse is the ratio of the distances from the centre
of the ellipse to one of the foci and to one of the vertices of the ellipse
✘ Formula of Eccentricity:-e=c/a
✘ Range of Eccentricity:- 0<e<1
9. 9
Questions
1)Find the coordinates of the foci, the vertices, the length of major axis, the minor axis, the eccentricity and the
length of the latus rectum of the ellipse.
1. x²/36 + y²/16 = 1
Solution:
The equation is x²/36 + y²/16 = 1.
On comparing the given equation with x2/a2 + y2/b2 = 1, we get
a = 6 and b = 4.
c = √(a2 – b2)= √(36-16)= √20= 2√5
Then,
The coordinates of the foci are (2√5, 0) and (-2√5, 0).
The coordinates of the vertices are (6, 0) and (-6, 0)
Length of major axis = 2a = 2 (6) = 12 & Length of minor axis = 2b = 2 (4) = 8
Eccentricity, e = c/a = 2√5/6 = √5/3
Length of latus rectum = 2b2/a = (2×16)/6 = 16/3
10. 10
Questions
2.Find the equation for the ellipse that satisfies the given conditions:
Vertices (± 5, 0), foci (± 4, 0)
Solution:
Here, the vertices are on the x-axis.
So, the equation of the ellipse will be of the form x²/a² + y²/b² = 1
Then, a = 5 and c = 4.
It is known that a2
= b² + c².
So, 5² = b² + 4²
25 = b² + 16
b² = 25 – 16
b = √9
= 3
∴ The equation of the ellipse is x²/52 + y²/32 = 1 or x²/25 + y²/9 = 1
11. 11
Questions
3.An arch is in the form of a semi-ellipse. It is 8 m wide and 2 m high at the centre. Find the height of the arch at
a point 1.5 m from one end.
Solution:
Since, the height and width of the arc from the centre is 2m and 8m respectively, it is clear that the length of the major axis is 8m,
while the length of the semi- minor axis is 2m.
The equation of the semi – ellipse will be of the from x2/16 + y2/4 = 1, y ≥ 0 … (1)
Let A be a point on the major axis such that AB = 1.5m.
Now draw AC ⊥ OB.
OA = (4 – 1.5)m = 2.5m i.e the x – coordinate of point C is 2.5
On substituting the value of x with 2.5 in equation (1), we get,
(2.5)2/16 + y2/4 = 1
6.25/16 + y2/4 = 1
y2 = 4 (1 – 6.25/16)= 4 (9.75/16)= 2.4375
y = 1.56 (approx.) i.e AC = 1.56m
Hence, the height of the arch at a point 1.5m from one end is approximately 1.56m
12. 12
Formula chart
✘ The standard form of the equation of an ellipse with center(0,0) is
where
✘ a>b
✘ the length of the major axis is 2a
✘ the length of minor axis is 2b
✘ the coordinates of vertices are (±a,0)
✘ the coordinates of foci are (±c,0)
✘ c=√a²-b²
✘ l=2b²/a
✘ e=c/a
