Class 11 Maths Ncert Solutions Chapter 11 Ex 11.3 Conic Sections PDF

# Class 11 Maths Ncert Solutions Ex 11.3

## Class 11 Maths Ncert Solutions Chapter 11 Ex 11.3

Q.1: Obtain the coordinates of the vertices, the foci, the major and the minor axis, the length of the latus rectum and the eccentricity of the ellipse x225+y29=1$\frac{x^{2}}{25} + \frac{y^{2}}{9} = 1$.

Sol:

Given:

x225+y29=1$\frac{x^{2}}{25} + \frac{y^{2}}{9} = 1$ . . . . . . . . (1) is the equation of the ellipse.

As we know that the denominator of x225$\frac{x^{2}}{25}$ > the denominator of y29$\frac{y^{2}}{9}$

So, along x – axis consist the major axis and along the y – axis consists the minor axis.

x2m2+y2n2=1$\frac{x^{2}}{m^{2}} + \frac{y^{2}}{n^{2}} = 1$ . . . . . . . . . . . . . . (2)

Now, on comparing equation (1) and equation (2) we will get:

c=m2n2=259=16=4$c = \sqrt{m^{2} – n^{2}} = \sqrt{25 – 9} = \sqrt{16} = 4$

We get:

m = 5, n = 3

The vertices coordinates are (m, 0) and (- m, 0) = (5, 0), (- 5, 0)

The foci’s coordinates are (c, 0) and (- c, 0) = (4, 0) and (- 4, 0)

Length of the axis:

Major axis = 2m = 10

Minor axis = 2n = 6

Length of the latus rectum = 2n2m$\frac{2n^{2}}{m}$ = 2.325=185$\frac{2.3^{2}}{5} = \frac{18}{5}$

Eccentricity, e = cm$\frac{c}{m}$ = 45$\frac{4}{5}$

Q.2: Obtain the coordinates of the vertices, the foci, the major and the minor axis, the length of the latus rectum and the eccentricity of the ellipse x29+y216=1$\frac{x^{2}}{9} + \frac{y^{2}}{16} = 1$.

Sol:

Given:

x29+y216=1$\frac{x^{2}}{9} + \frac{y^{2}}{16} = 1$ . . . . . . . . . . (1) is the equation of the ellipse.

As we know that the denominator of x29$\frac{x^{2}}{9}$ < the denominator of y216$\frac{y^{2}}{16}$

So, along y – axis consist the major axis and along the x – axis consists the minor axis.

x2n2+y2m2=1$\frac{x^{2}}{n^{2}} + \frac{y^{2}}{m^{2}} = 1$ . . . . . . . . . . . . (2)

Now, on comparing equation (1) and equation (2) we will get:

c=m2n2=169=7$c = \sqrt{m^{2} – n^{2}} = \sqrt{16 – 9} = \sqrt{7}$

We get,:

m = 4, n = 3

The vertices coordinates are (0, m) and (0, – m) = (0, 4), (0, – 4)

The foci’s coordinates are (0, c) and (0, – c) = (7$\sqrt{7}$, 0) and (-7$\sqrt{7}$, 0)

Length of the axis:

Major axis = 2m = 8

Minor axis = 2n = 6

Length of the latus rectum = 2n2m$\frac{2n^{2}}{m}$ = 2.324=92$\frac{2.3^{2}}{4} = \frac{9}{2}$

Eccentricity, e = cm$\frac{c}{m}$ = 74$\frac{\sqrt{7}}{4}$

Q.3: Obtain the coordinates of the vertices, the foci, the major and the minor axis, the length of the latus rectum and the eccentricity of the ellipse x225+y24=1$\frac{x^{2}}{25} + \frac{y^{2}}{4} = 1$.

Sol:

Given:

x225+y24=1$\frac{x^{2}}{25} + \frac{y^{2}}{4} = 1$ . . . . . . . . . . (1) is the equation of the ellipse.

As we know that the denominator of x225$\frac{x^{2}}{25}$ > the denominator of y24$\frac{y^{2}}{4}$

So, along x – axis consist the major axis and along the y – axis consists the minor axis.

x2m2+y2n2=1$\frac{x^{2}}{m^{2}} + \frac{y^{2}}{n^{2}} = 1$ . . . . . . . . . . (2)

Now, on comparing equation (1) and equation (2) we will get:

c=m2n2=254=21$c = \sqrt{m^{2} – n^{2}} = \sqrt{25 – 4} = \sqrt{21}$

We get:

m = 5, n = 2

The vertices coordinates are (m, 0) and (- m, 0) = (5, 0), (- 5, 0)

The foci’s coordinates are (c, 0) and (- c, 0) = (21$\sqrt{21}$, 0) and (-21$\sqrt{21}$, 0)

Length of the axis:

Major axis = 2m = 10

Minor axis = 2n = 4

Length of the latus rectum = 2n2m=2.225=85$\frac{2n^{2}}{m}=\frac{2.2^{2}}{5} = \frac{8}{5}$

Eccentricity, e = cm$\frac{c}{m}$ = 215$\frac{\sqrt{21}}{5}$

Q.4: Obtain the coordinates of the vertices, the foci, the major and the minor axis, the length of the latus rectum and the eccentricity of the ellipse x236+y2121=1$\frac{x^{2}}{36} + \frac{y^{2}}{121} = 1$.

Sol:

Given:
x236+y2121=1$\frac{x^{2}}{36} + \frac{y^{2}}{121} = 1$ . . . . . . . . . . (1) is the equation of the ellipse.

As we know that the denominator of x236$\frac{x^{2}}{36}$ < the denominator of y2121$\frac{y^{2}}{121}$

So, along y – axis consist the major axis and along the x – axis consists the minor axis.

x2n2+y2m2=1$\frac{x^{2}}{n^{2}} + \frac{y^{2}}{m^{2}} = 1$ . . . . . . . . . . . . . . . (2)

Now, on comparing equation (1) and equation (2) we will get:

c=m2n2=12136=85$c = \sqrt{m^{2} – n^{2}} = \sqrt{121 – 36} = \sqrt{85}$

We get:

m = 11, n = 6

The vertices coordinates are (0, m) and (0, – m) = (0, 11), (0, – 11)

The foci’s coordinates are (0, c) and (0, – c) = (0, 85$\sqrt{85}$) and (0, –85$\sqrt{85}$)

Length of the axis:

Major axis = 2m = 22

Minor axis = 2n = 12

Length of the latus rectum = 2n2m$\frac{2n^{2}}{m}$ = 2.6211=7211$\frac{2.6^{2}}{11} = \frac{72}{11}$

Eccentricity, e = cm$\frac{c}{m}$ = 8511$\frac{\sqrt{85}}{11}$

Q.5: Obtain the coordinates of the vertices, the foci, the major and the minor axis, the length of the latus rectum and the eccentricity of the ellipse x264+y225=1$\frac{x^{2}}{64} + \frac{y^{2}}{25} = 1$.

Sol:

Given:

x264+y225=1$\frac{x^{2}}{64} + \frac{y^{2}}{25} = 1$ . . . . . . . . . . . . . (1) is the equation of the ellipse.

As we know that the denominator of x264$\frac{x^{2}}{64}$ > the denominator of y225$\frac{y^{2}}{25}$

So, along x – axis consist the major axis and along the y – axis consists the minor axis.

x2m2+y2n2=1$\frac{x^{2}}{m^{2}} + \frac{y^{2}}{n^{2}} = 1$ . . . . . . . . . . . . . (2)

Now, on comparing equation (1) and equation (2) we will get:

c=m2n2=6425=39$c = \sqrt{m^{2} – n^{2}} = \sqrt{64 – 25} = \sqrt{39}$

We get:

m = 8, n = 5

The vertices coordinates are (m, 0) and (- m, 0) = (8, 0), (- 8, 0)

The foci’s coordinates are (c, 0) and (- c, 0) = (39$\sqrt{39}$, 0) and (-39$\sqrt{39}$, 0)

Length of the axis:

Major axis = 2m = 16

Minor axis = 2n = 10

Length of the latus rectum = 2n2m$\frac{2n^{2}}{m}$ = 2.528=254$\frac{2.5^{2}}{8} = \frac{25}{4}$

Eccentricity, e = cm$\frac{c}{m}$ = 398$\frac{\sqrt{39}}{8}$

Q.6: Obtain the coordinates of the vertices, the foci, the major and the minor axis, the length of the latus rectum and the eccentricity of the ellipse x2400+y2100=1$\frac{x^{2}}{400} + \frac{y^{2}}{100} = 1$.

Sol:

Given:

x2400+y2100=1$\frac{x^{2}}{400} + \frac{y^{2}}{100} = 1$ . . . . . . . . . . . . (1) is the equation of the ellipse.

As we know that the denominator of x2400$\frac{x^{2}}{400}$ > the denominator of y2100$\frac{y^{2}}{100}$

So, along x – axis consist the major axis and along the y – axis consists the minor axis.

x2m2+y2n2=1$\frac{x^{2}}{m^{2}} + \frac{y^{2}}{n^{2}} = 1$ . . . . . . . . . . . (2)

Now, on comparing equation (1) and equation (2) we will get:

c=m2n2=400100=300=103$c = \sqrt{m^{2} – n^{2}} = \sqrt{400 – 100} = \sqrt{300} = 10\sqrt{3}$

We get:

m = 20, n = 10

The vertices coordinates are (m, 0) and (- m, 0) = (20, 0), (- 20, 0)

The foci’s coordinates are (c, 0) and (- c, 0) = (103$10 \sqrt{3}$, 0) and (-103$10 \sqrt{3}$, 0)

Length of the axis:

Major axis = 2m = 40

Minor axis = 2n = 20

Length of the latus rectum = 2n2m$\frac{2n^{2}}{m}$ = 2.10220=10$\frac{2.10^{2}}{20} = 10$

Eccentricity, e = cm$\frac{c}{m}$ = 10320=32$\frac{10 \sqrt{3}}{20} = \frac{\sqrt{3}}{2}$

Q.7: Obtain the coordinates of the vertices, the foci, the major and the minor axis, the length of the latus rectum and the eccentricity of the ellipse 49x2 + 9y2 = 441

Sol:

Given:

49x2 + 9y2 = 441

x29+y249=1$\frac{x^{2}}{9} + \frac{y^{2}}{49} = 1$  . . . . . . . . . . . . . . . (1) is the equation of the ellipse.

As we know that the denominator of x29$\frac{x^{2}}{9}$ < the denominator of y249$\frac{y^{2}}{49}$

So, along y – axis consist the major axis and along the x – axis consists the minor axis.

x2n2+y2m2=1$\frac{x^{2}}{n^{2}} + \frac{y^{2}}{m^{2}} = 1$ . . . . . . . . . . . . . . . (2)

Now, on comparing equation (1) and equation (2) we will get:

c=m2n2=499=40=210$c = \sqrt{m^{2} – n^{2}} = \sqrt{49 – 9} = \sqrt{40} = 2 \sqrt{10}$

We get:

m = 7, n = 3

The vertices coordinates are (0, m) and (0, – m) = (0, 7), (0, – 7)

The foci’s coordinates are (0, c) and (0, – c) = (0, 210$2 \sqrt{10}$) and (0, 210$2 \sqrt{10}$)

Length of the axis:

Major axis = 2m = 14

Minor axis = 2n = 6

Length of the latus rectum = 2n2m$\frac{2n^{2}}{m}$ = 2.327=187$\frac{2.3^{2}}{7} = \frac{18}{7}$

Eccentricity, e = cm$\frac{c}{m}$ = 2107$\frac{2 \sqrt{10}}{7}$

Q.8: Obtain the coordinates of the vertices, the foci, the major and the minor axis, the length of the latus rectum and the eccentricity of the ellipse 4x2 + 16y2 = 64

Sol:

Given:

4x2 + 16y2 = 64

x216+y24=169$\frac{x^{2}}{16} + \frac{y^{2}}{4} = 169$ . . . . . . . . . . . . . . . . . (1) is the equation of the ellipse.

As we know that the denominator of x216$\frac{x^{2}}{16}$ > the denominator of y24$\frac{y^{2}}{4}$

So, along x – axis consist the major axis and along the y – axis consists the minor axis.

x2m2+y2n2=1$\frac{x^{2}}{m^{2}} + \frac{y^{2}}{n^{2}} = 1$ . . . . . . . . . . . . . . . (2)

Now, on comparing equation (1) and equation (2) we will get:

c=m2n2=164=12=23$c=\sqrt{m^{2}-n^{2}}=\sqrt{16 – 4}=\sqrt{12}=2\sqrt{3}$

We get:

m = 4, n = 2

The vertices coordinates are (m, 0) and (- m, 0) = (4, 0), (- 4, 0)

The foci’s coordinates are (c, 0) and (- c, 0) = (23$2 \sqrt{3}$, 0) and (-23$2 \sqrt{3}$, 0)

Length of the axis:

Major axis = 2m = 8

Minor axis = 2n = 4

Length of the latus rectum = 2n2m$\frac{2n^{2}}{m}$ = 2.424=8$\frac{2.4^{2}}{4} = 8$

Eccentricity, e = cm$\frac{c}{m}$ = 234=32$\frac{2 \sqrt{3}}{4} = \frac{\sqrt{3}}{2}$

Q.9: Obtain the coordinates of the vertices, the foci, the major and the minor axis, the length of the latus rectum and the eccentricity of the ellipse 4x2 + y2 = 4

Sol:

Given:

4x2 + y2 = 4

x21+y24=1$\frac{x^{2}}{1} + \frac{y^{2}}{4} = 1$ . . . . . . . . . . . . . . . . (1) is the equation of the ellipse.

As we know that the denominator of x21$\frac{x^{2}}{1}$ < the denominator of y24$\frac{y^{2}}{4}$

So, along y – axis consist the major axis and along the x – axis consists the minor axis.

x2n2+y2m2=1$\frac{x^{2}}{n^{2}} + \frac{y^{2}}{m^{2}} = 1$ . . . . . . . . . . . . . (2)

Now, on comparing equation (1) and equation (2) we will get:

c=m2n2=41=3$c = \sqrt{m^{2} – n^{2}} = \sqrt{4 – 1} = \sqrt{3}$

We get:

m = 2, n = 1

The vertices coordinates are (0, m) and (0, – m) = (0, 2), (0, – 2)

The foci’s coordinates are (0, c) and (0, – c) = (0, 3$\sqrt{3}$) and (0, –3$\sqrt{3}$)

Length of the axis:

Major axis = 2m = 4

Minor axis = 2n = 2

Length of the latus rectum = 2n2m$\frac{2n^{2}}{m}$ = 2.122=1$\frac{2.1^{2}}{2} = 1$

Eccentricity, e = cm$\frac{c}{m}$ = 32$\frac{\sqrt{3}}{2}$

Q.10: For the given condition obtain the equation of the ellipse.

(i) Vertices (±6, 0)

(ii) Foci (±3, 0)

Sol:

Given:

(i) Vertices (±6, 0)

(ii) Foci (±3, 0)

The vertices are represented along x – axis.

The required equation of the ellipse is of the form x2m2+y2n2=1$\frac{x^{2}}{m^{2}} + \frac{y^{2}}{n^{2}} = 1$ . . . . . . . . . . . . . . . . . (1)

We get:

m = 6 (semi major axis), c = 3

As we know:

m2 = n2 + c2

62 = n2 + 32

n2 = 62 – 32

n2 = 36 – 9 = 27

n = 27=33$\sqrt{27} = 3\sqrt{3}$

Hence, x236+y227=1$\frac{x^{2}}{36} + \frac{y^{2}}{27} = 1$  is the equation of the ellipse.

Q.11: For the given condition obtain the equation of the ellipse.

(i) Vertices (0, ±8)

(ii) Foci ( 0, ±4)

Sol:

Given:

(i) Vertices (0, ±8)

(ii) Foci ( 0, ±4)

The vertices are represented along y – axis.

The required equation of the ellipse is of the form  x2n2+y2m2=1$\frac{x^{2}}{n^{2}} + \frac{y^{2}}{m^{2}} = 1$ . . . . . . . . . . . . . . .(1)

We get:

m = 8 (semi major axis), c = 4

As we know:

m2 = n2 + c2

82 = n2 + 42

n2 = 82 – 42

n2 = 64 – 16 = 48

n = 48=43$\sqrt{48} = 4\sqrt{3}$

Hence, x248+y264=1$\frac{x^{2}}{48} + \frac{y^{2}}{64} = 1$ is the equation of the ellipse.

Q.12: For the given condition obtain the equation of the ellipse.

(i) Vertices (±5, 0)

(ii) Foci (±3, 0)

Sol:

Given:

(i) Vertices (± 5, 0)

(ii) Foci (± 3, 0)

The vertices are represented along x – axis.

The required equation of the ellipse is of the form  x2m2+y2n2=1$\frac{x^{2}}{m^{2}} + \frac{y^{2}}{n^{2}} = 1$ . . . . . . . . . . . . . . . (1)

We get:

m = 5 (semi major axis), c = 3

As we know:

m2 = n2 + c2

52 = n2 + 32

n2 = 52 – 32

n2 = 25 – 9 = 16

n = 16=4$\sqrt{16} = 4$

Hence, x225+y216=1$\frac{x^{2}}{25} + \frac{y^{2}}{16} = 1$ is the equation of the ellipse.

Q.13: For the given condition obtain the equation of the ellipse.

(i) Coordinates of major axis (±5, 0)

(ii) Coordinates of minor axis (0, ±3)

Sol:

Given:

(i) Coordinates of major axis (±5, 0)

(ii) Coordinates of minor axis (0, ±3)

The major axis is represented along x – axis.

The required equation of the ellipse is of the form

x2m2+y2n2=1$\frac{x^{2}}{m^{2}} + \frac{y^{2}}{n^{2}} = 1$ . . . . . . . . . . . . . . . (1)

We get:

m = 5 (semi major axis), n = 3

Hence, x225+y216=1$\frac{x^{2}}{25} + \frac{y^{2}}{16} = 1$ is the equation of the ellipse.

Q.14: For the given condition obtain the equation of the ellipse.

(i) Coordinates of major axis (0, ±2)

(ii) Coordinates of minor axis (±3$\sqrt{3}$, 0)

Sol:

Given:

(i) Coordinates of major axis (0, ±2)

(ii) Coordinates of minor axis (±3$\sqrt{3}$, 0)

The major axis is represented along y – axis.

The required equation of the ellipse is of the form  x2n2+y2m2=1$\frac{x^{2}}{n^{2}} + \frac{y^{2}}{m^{2}} = 1$ . . . . . . . . . . . . . . . . . (1)

We get:

m = 2 (semi major axis), n = 3$\sqrt{3}$

Hence, x23+y24=1$\frac{x^{2}}{3} + \frac{y^{2}}{4} = 1$ is the equation of the ellipse.

Q.15: For the given condition obtain the equation of the ellipse.

(i) Length of major axis 30

(ii) Coordinates of foci (±4, 0)

Sol:

Given:

(i) Length of major axis 30

(ii) Coordinates of foci (±4, 0)

The major axis are represented along x – axis as foci is along x – axis

The required equation of the ellipse is of the form x2m2+y2n2=1$\frac{x^{2}}{m^{2}} + \frac{y^{2}}{n^{2}} = 1$ . . . . . . . . . . . . . . . . . . . (1)

We get:

2m = 30 (semi major axis),

m = 15

c = 4

As we know:

m2 = n2 + c2

152 = n2 + 42

n2 = 152 – 42

n2 = 225 – 16 = 209

n = 209$\sqrt{209}$

Hence, x2225+y2209=1$\frac{x^{2}}{225} + \frac{y^{2}}{209} = 1$ is the equation of the ellipse.

Q.16: For the given condition obtain the equation of the ellipse.

(i) Length of minor axis 26

(ii) Coordinates of foci (0, ±9)

Sol:

Given:

(i) Length of minor axis 26

(ii) Coordinates of foci (0, ±9)

The major axis are represented along y – axis as foci is along y – axis

The required equation of the ellipse is of the form

x2n2+y2m2=1$\frac{x^{2}}{n^{2}} + \frac{y^{2}}{m^{2}} = 1$ . . . . . . . . . . . . . (1)

We get:

2n = 26 (semi major axis),

n = 13

c = 9

As we know:

m2 = n2 + c2

m2 = 132 + 92

m2 = 169 + 81

m2 = 250

m = 250$\sqrt{250}$ = 510$5 \sqrt{10}$

Hence, x2169+y2250=1$\frac{x^{2}}{169} + \frac{y^{2}}{250} = 1$ is the equation of the ellipse.

Q.17: For the given condition obtain the equation of the ellipse.

(i) Coordinates of foci (±4, 0)

(ii) m = 6

Sol:

Given:

(i) Coordinates of foci (±4, 0)

(ii) m = 6

The major axis are represented along x – axis as foci is along x – axis

The required equation of the ellipse is of the form

x2m2+y2n2=1$\frac{x^{2}}{m^{2}} + \frac{y^{2}}{n^{2}} = 1$ . . . . . . . . . . . . . . . (1)

We get:

c = 4

m = 6

As we know:

m2 = n2 + c2

62 = n2 + 42

n2 = 36 – 16

n2 = 20

n = 20$\sqrt{20}$ = 25$2 \sqrt{5}$

Hence, x236+y220=1$\frac{x^{2}}{36} + \frac{y^{2}}{20} = 1$ is the equation of the ellipse.

Q.18: For the given condition obtain the equation of the ellipse.

(i) n = 2, c = 3 (on the x – axis)

(ii) Coordinates of centre (0, 0)

Sol:

Given:

(i) n = 2, c = 3 (on the x – axis)

(ii) Coordinates of centre (0, 0) i.e., centre is at origin.

The major axis are represented along x – axis as foci is along x – axis

The required equation of the ellipse is of the form:

x2m2+y2n2=1$\frac{x^{2}}{m^{2}} + \frac{y^{2}}{n^{2}} = 1$ . . . . . . . . . . . . . (1)

We have:

n = 2, c = 3

As we know:

m2 = n2 + c2

m2 = 22 + 32

m2 = 4 + 9

m2 = 13

m = 13$\sqrt{13}$

Hence, x213+y24=1$\frac{x^{2}}{13} + \frac{y^{2}}{4} = 1$ is the equation of the ellipse.

Q.19: For the given condition obtain the equation of the ellipse.

(i) Centre is at origin and major axis is along the y – axis

(ii) It passes through the points (1, 6) and (3, 2)

Sol:

Given:

(i) Centre is at origin and major axis is along the y – axis

(ii) It passes through the points (1, 6) and (3, 2)

The major axis are represented along y – axis as foci is along y – axis

The required equation of the ellipse is of the form:

x2n2+y2m2=1$\frac{x^{2}}{n^{2}} + \frac{y^{2}}{m^{2}} = 1$ . . . . . . . . . . . . . . . (1)

As the ellipse passes through (2, 1) and (2, 3) points, then

12n2+62m2=1$\frac{1^{2}}{n^{2}} + \frac{6^{2}}{m^{2}} = 1$ 32n2+22m2=1$\frac{3^{2}}{n^{2}} + \frac{2^{2}}{m^{2}} = 1$

1n2+36m2$\frac{1}{n^{2}} + \frac{36}{m^{2}}$ = 1 . . . . . . . . . . . . . . . (a)

9n2+4m2$\frac{9}{n^{2}} + \frac{4}{m^{2}}$ = 1 . . . . . . . . . . . . . . . . (b)

Now, on comparing equation (a) and equation (b) we will get:

m2 = 40, n2 = 10

Hence, y240+x210=1$\frac{y^{2}}{40} + \frac{x^{2}}{10} = 1$ is the equation of the ellipse.

Q.20: For the given condition obtain the equation of the ellipse.

(i) Major axis is represented along x – axis

(ii) It passes through the points (6, 2) and (4, 3)

Sol:

Given:

(i) Major axis is represented along x – axis

(ii) It passes through the points (6, 2) and (4, 3)

The major axis are represented along y – axis as foci is along y – axis

The required equation of the ellipse is of the form:

x2m2+y2n2=1$\frac{x^{2}}{m^{2}} + \frac{y^{2}}{n^{2}} = 1$ . . . . . . . . . . . . . . . . (1)

As the ellipse passes through (6, 2) and (4, 3) points, then

62m2+22n2=1$\frac{6^{2}}{m^{2}} + \frac{2^{2}}{n^{2}} = 1$ 42m2+32n2=1$\frac{4^{2}}{m^{2}} + \frac{3^{2}}{n^{2}} = 1$

36m2+4n2$\frac{36}{m^{2}} + \frac{4}{n^{2}}$ = 1 . . . . . . . . . . . . . . . . (a)

16m2+9n2$\frac{16}{m^{2}} + \frac{9}{n^{2}}$ = 1 . . . . . . . . . . . . . . . . (b)

Now, on comparing equation (a) and equation (b) we will get:

m2 = 52 and n2 = 13

Hence, y213+x252=1$\frac{y^{2}}{13} + \frac{x^{2}}{52} = 1$ is the equation of the ellipse.