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An ellipse is the set of all points in a plane, the sum of whose distances from two fixed points in the plane is a constant. There are two axes of the ellipse, the major and minor axes. In this exercise, you will learn about the Standard equation of an ellipse, the Eccentricity of an ellipse, and the Latus rectum of an ellipse. Many real-world situations, like orbits of planets, orbits of satellites, orbits of moons, and some aeroplane wings, are represented by an ellipse. It also has many applications in the field of science and research. There are some definitions, theories, and observations related to the ellipse given in the NCERT before this exercise. You must go through these theories in order to get conceptual clarity.
The NCERT solutions of Chapter 10 conic section exercise 10.3 are designed by respective subject matter experts to offer a systematic approach to these important concepts and help students to prepare well for board exams by a series of solved questions, which are given in exercise 10.3. These NCERT Solutions also provide a valuable resource to the students to enhance their performance in their board exams as well as various competitive exams.
$\frac{x^2}{36} + \frac{y^2}{16} = 1$
Answer:
Given
The equation of the ellipse
$\frac{x^2}{36} + \frac{y^2}{16} = 1$
As we can see from the equation, the major axis is along X-axis and the minor axis is along Y-axis.
On comparing the given equation with the standard equation of an ellipse, which is
$\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$
We get
$a=6$ and $b=4$.
So,
$c=\sqrt{a^2-b^2}=\sqrt{6^2-4^2}$
$c=\sqrt{20}=2\sqrt{5}$
Hence,
Coordinates of the foci:
$(c,0)\:and\:(-c,0)=(2\sqrt{5},0)\:and\:(-2\sqrt{5},0)$
The vertices:
$(a,0)\:and\:(-a,0)=(6,0)\:and\:(-6,0)$
The length of the major axis:
$2a=2(6)=12$
The length of minor axis:
$2b=2(4)=8$
The eccentricity :
$e=\frac{c}{a}=\frac{2\sqrt{5}}{6}=\frac{\sqrt{5}}{3}$
The length of the latus rectum:
$\frac{2b^2}{a}=\frac{2(4)^2}{6}=\frac{32}{6}=\frac{16}{3}$
$\frac{x^2}{4} + \frac{y^2}{25} =1$
Answer:
Given
The equation of the ellipse
$\frac{x^2}{4} + \frac{y^2}{25} =1$
As we can see from the equation, the major axis is along Y-axis and the minor axis is along X-axis.
On comparing the given equation with the standard equation of such ellipse, which is
$\frac{x^2}{b^2} + \frac{y^2}{a^2} = 1$
We get
$a=5$ and $b=2$.
So,
$c=\sqrt{a^2-b^2}=\sqrt{5^2-2^2}$
$c=\sqrt{21}$
Hence,
Coordinates of the foci:
$(0,c)\:and\:(0,-c)=(0,\sqrt{21})\:and\:(0,-\sqrt{21})$
The vertices:
$(0,a)\:and\:(0,-a)=(0,5)\:and\:(0,-5)$
The length of the major axis:
$2a=2(5)=10$
The length of minor axis:
$2b=2(2)=4$
The eccentricity :
$e=\frac{c}{a}=\frac{\sqrt{21}}{6}$
The length of the latus rectum:
$\frac{2b^2}{a}=\frac{2(2)^2}{5}=\frac{8}{5}$
$\frac{x^2}{16} + \frac{y^2}{9} = 1$
Answer:
Given
The equation of the ellipse
$\frac{x^2}{16} + \frac{y^2}{9} = 1$
As we can see from the equation, the major axis is along X-axis and the minor axis is along Y-axis.
On comparing the given equation with the standard equation of an ellipse, which is
$\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$
We get
$a=4$ and $b=3$.
So,
$c=\sqrt{a^2-b^2}=\sqrt{4^2-3^2}$
$c=\sqrt{7}$
Hence,
Coordinates of the foci:
$(c,0)\:and\:(-c,0)=(\sqrt{7},0)\:and\:(-\sqrt{7},0)$
The vertices:
$(a,0)\:and\:(-a,0)=(4,0)\:and\:(-4,0)$
The length of the major axis:
$2a=2(4)=8$
The length of minor axis:
$2b=2(3)=6$
The eccentricity :
$e=\frac{c}{a}=\frac{\sqrt{7}}{4}$
The length of the latus rectum:
$\frac{2b^2}{a}=\frac{2(3)^2}{4}=\frac{18}{4}=\frac{9}{2}$
$\frac{x^2}{25} + \frac{y^2}{100} = 1$
Answer:
Given
The equation of the ellipse
$\frac{x^2}{25} + \frac{y^2}{100} = 1$
As we can see from the equation, the major axis is along Y-axis and the minor axis is along X-axis.
On comparing the given equation with the standard equation of such ellipse, which is
$\frac{x^2}{b^2} + \frac{y^2}{a^2} = 1$
We get
$a=10$ and $b=5$.
So,
$c=\sqrt{a^2-b^2}=\sqrt{10^2-5^2}$
$c=\sqrt{75}=5\sqrt{3}$
Hence,
Coordinates of the foci:
$(0,c)\:and\:(0,-c)=(0,5\sqrt{3})\:and\:(0,-5\sqrt{3})$
The vertices:
$(0,a)\:and\:(0,-a)=(0,10)\:and\:(0,-10)$
The length of the major axis:
$2a=2(10)=20$
The length of minor axis:
$2b=2(5)=10$
The eccentricity :
$e=\frac{c}{a}=\frac{5\sqrt{3}}{10}=\frac{\sqrt{3}}{2}$
The length of the latus rectum:
$\frac{2b^2}{a}=\frac{2(5)^2}{10}=\frac{50}{10}=5$
$\frac{x^2}{49} + \frac{y^2}{36} = 1$
Answer:
Given
The equation of ellipse
$\frac{x^2}{49} + \frac{y^2}{36} = 1\Rightarrow \frac{x^2}{7^2} + \frac{y^2}{6^2} = 1$
As we can see from the equation, the major axis is along X-axis and the minor axis is along Y-axis.
On comparing the given equation with standard equation of ellipse, which is
$\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$
We get
$a=7$ and $b=6$.
So,
$c=\sqrt{a^2-b^2}=\sqrt{7^2-6^2}$
$c=\sqrt{13}$
Hence,
Coordinates of the foci:
$(c,0)\:and\:(-c,0)=(\sqrt{13},0)\:and\:(-\sqrt{13},0)$
The vertices:
$(a,0)\:and\:(-a,0)=(7,0)\:and\:(-7,0)$
The length of major axis:
$2a=2(7)=14$
The length of minor axis:
$2b=2(6)=12$
The eccentricity :
$e=\frac{c}{a}=\frac{\sqrt{13}}{7}$
The length of the latus rectum:
$\frac{2b^2}{a}=\frac{2(6)^2}{7}=\frac{72}{7}$
$\frac{x^2}{100} + \frac{y^2}{400} =1$
Answer:
Given
The equation of the ellipse
$\frac{x^2}{100} + \frac{y^2}{400} =1\Rightarrow \frac{x^2}{10^2} + \frac{y^2}{20^2} =1$
As we can see from the equation, the major axis is along Y-axis and the minor axis is along X-axis.
On comparing the given equation with the standard equation of such ellipse, which is
$\frac{x^2}{b^2} + \frac{y^2}{a^2} = 1$
We get
$a=20$ and $b=10$.
So,
$c=\sqrt{a^2-b^2}=\sqrt{20^2-10^2}$
$c=\sqrt{300}=10\sqrt{3}$
Hence,
Coordinates of the foci:
$(0,c)\:and\:(0,-c)=(0,10\sqrt{3})\:and\:(0,-10\sqrt{3})$
The vertices:
$(0,a)\:and\:(0,-a)=(0,20)\:and\:(0,-20)$
The length of the major axis:
$2a=2(20)=40$
The length of minor axis:
$2b=2(10)=20$
The eccentricity :
$e=\frac{c}{a}=\frac{10\sqrt{3}}{20}=\frac{\sqrt{3}}{2}$
The length of the latus rectum:
$\frac{2b^2}{a}=\frac{2(10)^2}{20}=\frac{200}{20}=10$
Answer:
Given
The equation of the ellipse
$36x^2 + 4y^2 =144$
$\Rightarrow \frac{36}{144}x^2 + \frac{4}{144}y^2 =1$
$\Rightarrow \frac{1}{4}x^2 + \frac{1}{36}y^2 =1$
$\frac{x^2}{2^2} + \frac{y^2}{6^2} = 1$
As we can see from the equation, the major axis is along Y-axis and the minor axis is along X-axis.
On comparing the given equation with the standard equation of such ellipse, which is
$\frac{x^2}{b^2} + \frac{y^2}{a^2} = 1$
We get
$a=6$ and $b=2$.
So,
$c=\sqrt{a^2-b^2}=\sqrt{6^2-2^2}$
$c=\sqrt{32}=4\sqrt{2}$
Hence,
Coordinates of the foci:
$(0,c)\:and\:(0,-c)=(0,4\sqrt{2})\:and\:(0,-4\sqrt{2})$
The vertices:
$(0,a)\:and\:(0,-a)=(0,6)\:and\:(0,-6)$
The length of the major axis:
$2a=2(6)=12$
The length of minor axis:
$2b=2(2)=4$
The eccentricity :
$e=\frac{c}{a}=\frac{4\sqrt{2}}{6}=\frac{2\sqrt{2}}{3}$
The length of the latus rectum:
$\frac{2b^2}{a}=\frac{2(2)^2}{6}=\frac{8}{6}=\frac{4}{3}$
Answer:
Given
The equation of the ellipse
$16x^2 + y^2 = 16$
$\frac{16x^2}{16} + \frac{y^2}{16} = 1$
$\frac{x^2}{1^2} + \frac{y^2}{4^2} = 1$
As we can see from the equation, the major axis is along Y-axis and the minor axis is along X-axis.
On comparing the given equation with the standard equation of such ellipse, which is
$\frac{x^2}{b^2} + \frac{y^2}{a^2} = 1$
We get
$a=4$ and $b=1$.
So,
$c=\sqrt{a^2-b^2}=\sqrt{4^2-1^2}$
$c=\sqrt{15}$
Hence,
Coordinates of the foci:
$(0,c)\:and\:(0,-c)=(0,\sqrt{15})\:and\:(0,-\sqrt{15})$
The vertices:
$(0,a)\:and\:(0,-a)=(0,4)\:and\:(0,-4)$
The length of the major axis:
$2a=2(4)=8$
The length of minor axis:
$2b=2(1)=2$
The eccentricity :
$e=\frac{c}{a}=\frac{\sqrt{15}}{4}$
The length of the latus rectum:
$\frac{2b^2}{a}=\frac{2(1)^2}{4}=\frac{2}{4}=\frac{1}{2}$
Answer:
Given
The equation of the ellipse
$4x^2 + 9y^2 =36$
$\frac{4x^2}{36} + \frac{9y^2}{36} = 1$
$\frac{x^2}{9} + \frac{y^2}{4} = 1$
$\frac{x^2}{3^2} + \frac{y^2}{2^2} = 1$
As we can see from the equation, the major axis is along X-axis and the minor axis is along Y-axis.
On comparing the given equation with the standard equation of an ellipse, which is
$\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$
We get
$a=3$ and $b=2$.
So,
$c=\sqrt{a^2-b^2}=\sqrt{3^2-2^2}$
$c=\sqrt{5}$
Hence,
Coordinates of the foci:
$(c,0)\:and\:(-c,0)=(\sqrt{5},0)\:and\:(-\sqrt{5},0)$
The vertices:
$(a,0)\:and\:(-a,0)=(3,0)\:and\:(-3,0)$
The length of the major axis:
$2a=2(3)=6$
The length of minor axis:
$2b=2(2)=4$
The eccentricity :
$e=\frac{c}{a}=\frac{\sqrt{5}}{3}$
The length of the latus rectum:
$\frac{2b^2}{a}=\frac{2(2)^2}{3}=\frac{8}{3}$
Question 10: Find the equation for the ellipse that satisfies the given conditions:
Vertices (± 5, 0), foci (± 4, 0)
Answer:
Given, In an ellipse,
Vertices (± 5, 0), foci (± 4, 0)
Here Vertices and focus of the ellipse are in X-axis so the major axis of this ellipse will be X-axis.
Therefore, the equation of the ellipse will be of the form:
$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$
Where $a$ and $b$ are the length of the semimajor axis and semiminor axis respectively.
So on comparing standard parameters( vertices and foci) with the given one, we get
$a=5$ and $c=4$
Now, As we know the relation,
$a^2=b^2+c^2$
$b^2=a^2-c^2$
$b=\sqrt{a^2-c^2}$
$b=\sqrt{5^2-4^2}$
$b=\sqrt{9}$
$b=3$
Hence, The Equation of the ellipse will be :
$\frac{x^2}{5^2}+\frac{y^2}{3^2}=1$
Which is
$\frac{x^2}{25}+\frac{y^2}{9}=1$.
Question 11: Find the equation for the ellipse that satisfies the given conditions:
Vertices (0, ± 13), foci (0, ± 5)
Answer:
Given, In an ellipse,
Vertices (0, ± 13), foci (0, ± 5)
Here Vertices and focus of the ellipse are in Y-axis so the major axis of this ellipse will be Y-axis.
Therefore, the equation of the ellipse will be of the form:
$\frac{x^2}{b^2}+\frac{y^2}{a^2}=1$
Where $a$ and $b$are the length of the semimajor axis and semiminor axis respectively.
So on comparing standard parameters( vertices and foci) with the given one, we get
$a=13$ and $c=5$
Now, As we know the relation,
$a^2=b^2+c^2$
$b^2=a^2-c^2$
$b=\sqrt{a^2-c^2}$
$b=\sqrt{13^2-5^2}$
$b=\sqrt{169-25}$
$b=\sqrt{144}$
$b=12$
Hence, The Equation of the ellipse will be :
$\frac{x^2}{12^2}+\frac{y^2}{13^3}=1$
Which is
$\frac{x^2}{144}+\frac{y^2}{169}=1$.
Question 12: Find the equation for the ellipse that satisfies the given conditions:
Vertices (± 6, 0), foci (± 4, 0)
Answer:
Given, In an ellipse,
Vertices (± 6, 0), foci (± 4, 0)
Here Vertices and focus of the ellipse are in X-axis so the major axis of this ellipse will be X-axis.
Therefore, the equation of the ellipse will be of the form:
$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$
Where $a$ and $b$are the length of the semimajor axis and semiminor axis respectively.
So on comparing standard parameters( vertices and foci) with the given one, we get
$a=6$ and $c=4$
Now, As we know the relation,
$a^2=b^2+c^2$
$b^2=a^2-c^2$
$b=\sqrt{a^2-c^2}$
$b=\sqrt{6^2-4^2}$
$b=\sqrt{36-16}$
$b=\sqrt{20}$
Hence, The Equation of the ellipse will be :
$\frac{x^2}{6^2}+\frac{y^2}{(\sqrt{20})^2}=1$
Which is
$\frac{x^2}{36}+\frac{y^2}{20}=1$.
Question 13: Find the equation for the ellipse that satisfies the given conditions:
Ends of major axis (± 3, 0), ends of minor axis (0, ± 2)
Answer:
Given, In an ellipse,
Ends of the major axis (± 3, 0), ends of minor axis (0, ± 2)
Here, the major axis of this ellipse will be X-axis.
Therefore, the equation of the ellipse will be of the form:
$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$
Where $a$ and $b$are the length of the semimajor axis and semiminor axis respectively.
So on comparing standard parameters( ends of the major and minor axis ) with the given one, we get
$a=3$ and $b=2$
Hence, The Equation of the ellipse will be :
$\frac{x^2}{3^2}+\frac{y^2}{2^2}=1$
Which is
$\frac{x^2}{9}+\frac{y^2}{4}=1$.
Question 14: Find the equation for the ellipse that satisfies the given conditions:
Ends of major axis (0, ± $\sqrt{5}$ ), ends of minor axis (± 1, 0)
Answer:
Given, In an ellipse,
Ends of the major axis (0, ±$\sqrt{5}$ ), ends of minor axis (± 1, 0)
Here, the major axis of this ellipse will be Y-axis.
Therefore, the equation of the ellipse will be of the form:
$\frac{x^2}{b^2}+\frac{y^2}{a^2}=1$
Where $a$ and $b$are the length of the semimajor axis and semiminor axis respectively.
So on comparing standard parameters( ends of the major and minor axis ) with the given one, we get
$a=\sqrt{5}$ and $b=1$
Hence, The Equation of the ellipse will be :
$\frac{x^2}{1^2}+\frac{y^2}{(\sqrt{5})^2}=1$
Which is
$\frac{x^2}{1}+\frac{y^2}{5}=1$.
Question 15: Find the equation for the ellipse that satisfies the given conditions:
Length of major axis 26, foci (± 5, 0)
Answer:
Given, In an ellipse,
Length of major axis 26, foci (± 5, 0)
Here, the focus of the ellipse is in X-axis so the major axis of this ellipse will be X-axis.
Therefore, the equation of the ellipse will be of the form:
$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$
Where $a$ and $b$are the length of the semimajor axis and semiminor axis respectively.
So on comparing standard parameters( Length of semimajor axis and foci) with the given one, we get
$2a=26\Rightarrow a=13$ and $c=5$
Now, As we know the relation,
$a^2=b^2+c^2$
$b^2=a^2-c^2$
$b=\sqrt{a^2-c^2}$
$b=\sqrt{13^2-5^2}$
$b=\sqrt{144}$
$b=12$
Hence, The Equation of the ellipse will be :
$\frac{x^2}{13^2}+\frac{y^2}{12^2}=1$
Which is
$\frac{x^2}{169}+\frac{y^2}{144}=1$.
Question 16: Find the equation for the ellipse that satisfies the given conditions:
Length of minor axis 16, foci (0, ± 6).
Answer:
Given, In an ellipse,
Length of minor axis 16, foci (0, ± 6).
Here, the focus of the ellipse is on the Y-axis so the major axis of this ellipse will be Y-axis.
Therefore, the equation of the ellipse will be of the form:
$\frac{x^2}{b^2}+\frac{y^2}{a^2}=1$
Where $a$ and $b$are the length of the semimajor axis and semiminor axis respectively.
So on comparing standard parameters( length of semi-minor axis and foci) with the given one, we get
$2b=16\Rightarrow b=8$ and $c=6$
Now, As we know the relation,
$a^2=b^2+c^2$
$a=\sqrt{b^2+c^2}$
$a=\sqrt{8^2+6^2}$
$a=\sqrt{64+36}$
$a=\sqrt{100}$
$a=10$
Hence, The Equation of the ellipse will be :
$\frac{x^2}{8^2}+\frac{y^2}{10^3}=1$
Which is
$\frac{x^2}{64}+\frac{y^2}{100}=1$.
Question 17: Find the equation for the ellipse that satisfies the given conditions:
Foci (± 3, 0), a = 4
Answer:
Given, In an ellipse,
V Foci (± 3, 0), a = 4
Here foci of the ellipse are in X-axis so the major axis of this ellipse will be X-axis.
Therefore, the equation of the ellipse will be of the form:
$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$
Where $a$ and $b$are the length of the semimajor axis and semiminor axis respectively.
So on comparing standard parameters( vertices and foci) with the given one, we get
$a=4$ and $c=3$
Now, As we know the relation,
$a^2=b^2+c^2$
$b^2=a^2-c^2$
$b=\sqrt{a^2-c^2}$
$b=\sqrt{4^2-3^2}$
$b=\sqrt{7}$
Hence, The Equation of the ellipse will be :
$\frac{x^2}{4^2}+\frac{y^2}{(\sqrt{7})^2}=1$
Which is
$\frac{x^2}{16}+\frac{y^2}{7}=1$.
Question 18: Find the equation for the ellipse that satisfies the given conditions:
b = 3, c = 4, centre at the origin; foci on the x axis.
Answer:
Given,In an ellipse,
b = 3, c = 4, centre at the origin; foci on the x axis.
Here foci of the ellipse are in X-axis so the major axis of this ellipse will be X-axis.
Therefore, the equation of the ellipse will be of the form:
$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$
Where $a$ and $b$are the length of the semimajor axis and semiminor axis respectively.
Also Given,
$b=3$ and $c=4$
Now, As we know the relation,
$a^2=b^2+c^2$
$a^2=3^2+4^2$
$a^2=25$
$a=5$
Hence, The Equation of the ellipse will be :
$\frac{x^2}{5^2}+\frac{y^2}{3^2}=1$
Which is
$\frac{x^2}{25}+\frac{y^2}{9}=1$.
Question 19: Find the equation for the ellipse that satisfies the given conditions:
Centre at (0,0), major axis on the y-axis and passes through the points (3, 2) and (1,6).
Answer:
Given,in an ellipse
Centre at (0,0), major axis on the y-axis and passes through the points (3, 2) and (1,6).
Since, The major axis of this ellipse is on the Y-axis, the equation of the ellipse will be of the form:
$\frac{x^2}{b^2}+\frac{y^2}{a^2}=1$
Where $a$ and $b$ are the length of the semimajor axis and semiminor axis respectively.
Now since the ellipse passes through points,(3, 2)
$\frac{3^2}{b^2}+\frac{2^2}{a^2}=1$
${9a^2+4b^2}={a^2b^2}$
since the ellipse also passes through points,(1, 6).
$\frac{1^2}{b^2}+\frac{6^2}{a^2}=1$
$a^2+36b^2=a^2b^2$
On solving these two equation we get
$a^2=40$ and $b^2=10$
Thus, The equation of the ellipse will be
$\frac{x^2}{10}+\frac{y^2}{40}=1$.
Question 20: Find the equation for the ellipse that satisfies the given conditions:
Major axis on the x-axis and passes through the points (4,3) and (6,2).
Answer:
Given, in an ellipse
Major axis on the x-axis and passes through the points (4,3) and (6,2).
Since The major axis of this ellipse is on the X-axis, the equation of the ellipse will be of the form:
$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$
Where $a$ and $b$are the length of the semimajor axis and semiminor axis respectively.
Now since the ellipse passes through the point,(4,3)
$\frac{4^2}{a^2}+\frac{3^2}{b^2}=1$
${16b^2+9a^2}={a^2b^2}$
since the ellipse also passes through the point (6,2).
$\frac{6^2}{a^2}+\frac{2^2}{b^2}=1$
$4a^2+36b^2=a^2b^2$
On solving this two equation we get
$a^2=52$ and $b^2=13$
Thus, The equation of the ellipse will be
$\frac{x^2}{52}+\frac{y^2}{13}=1$
Also Read
1) What is an Ellipse
An ellipse is the set of all points in a plane whose distances from two fixed points are constant. And those fixed points are foci and the plural of focus.
2) Eccentricity of an Ellipse
The eccentricity of an ellipse tells us how oval-shaped the ellipse is.
The eccentricity (e) of an ellipse is given by:
$ e=\frac{c}{a}$ where, c = $\sqrt{a^{2}-b^{2}}$
$e=\frac{\sqrt{a^{2}-b^{2}}}{a}$
Where:
a = half the longest diameter (semi-major axis)
c = distance from the centre to each focus
b = half the shortest diameter (semi-minor axis)
also, 0 < e < 1 for an ellipse.
3) Standard equations of an ellipse: An ellipse has two standard forms, based on the direction of its major axis.
Horizontal Major Axis:
$\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}= 1$
Where, a > b (major axis is horizontal), Centre: (0,0)
Vertical Major Axis:
$\frac{x^{2}}{b^{2}}+\frac{y^{2}}{a^{2}}= 1$
Where, a < b (major axis is vertical), Centre: (0,0)
4) Latus rectum of an Ellipse: It is a line segment that passes through one of the ellipse's foci and is perpendicular to the major axis.
Length of the latus rectum of an ellipse for the standard equation
$=\frac{2b^{2}}{a}$
Note: Latus rectum is the same for both Horizontal Major Axis and Vertical Major Axis.
Also Read
Students can refer subject wise NCERT solutions. The links to solutions are given below
Students can access the NCERT exemplar solutions to enhance their deep understanding of the topic. These solutions are aligned with the CBSE syllabus and also help in competitive exams.
The set of all points in a plane, the sum of whose distances from two fixed points in the plane is a constant is called an ellipse.
The centre of the ellipse is the mid-point of the line segment joining the foci.
The major axis of an ellipse is a line segment passing through the foci of the ellipse.
The minor axis of an ellipse is a line segment passing through the centre and perpendicular to the major axis.
The Latus rectum of an ellipse is a line segment passing through the foci and perpendicular to the major axis whose endpoints lie on the ellipse.
The whole coordinate geometry unit has 10 marks weightage in the CBSE Class 11 Maths final exam.
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