GSEB Solutions Class 11 Maths Chapter 11 Conic Sections Ex 11.3

   

Gujarat Board GSEB Textbook Solutions Class 11 Maths Chapter 11 Conic Sections Ex 11.3 Textbook Questions and Answers.

Gujarat Board Textbook Solutions Class 11 Maths Chapter 11 Conic Sections Ex 11.3

In each of the following questions 1 to 9, find the co-ordinates of the foci, the vertices, the length of major axis, the minor axis, the eccentricity and length of latus rectum of the ellipse:
1. \(\frac{x^{2}}{36}\) + \(\frac{y^{2}}{16}\) = 1
2. \(\frac{x^{2}}{4}\) + \(\frac{y^{2}}{25}\) = 1
3. \(\frac{x^{2}}{16}\) + \(\frac{y^{2}}{9}\) = 1
4. \(\frac{x^{2}}{25}\) + \(\frac{y^{2}}{100}\) = 1
5. \(\frac{x^{2}}{49}\) + \(\frac{y^{2}}{36}\) = 1
6. \(\frac{x^{2}}{100}\) + \(\frac{y^{2}}{400}\) = 1
7. 36x2 + 4y2 = 144
8. 16x2 + y2 = 16
9. 4x2 + 9y2 = 36.
Solutions to questions 1 to 9:
1. Equation of ellipse is
\(\frac{x^{2}}{36}\) + \(\frac{y^{2}}{16}\) = 1.
Here, a2 = 36, b2 = 16.
∴ a = 6, b = 4
c2 = a2 – b2 = 36 – 16 = 20.
c = ± \(\sqrt{20}\), = ±2 \(\sqrt{5}\).
c = ae.
∴ e = \(\frac{c}{a}\) = \(\frac{2 \sqrt{5}}{6}\) = \(\frac{\sqrt{5}}{3}\)
Co-ordinates of foci are (± c, 0) i.e., (± 2\(\sqrt{5}\), 0).
Vertices are (± a, 0) i.e., (± 6, 0).
Length of major axis = 2a = 2 × 6 = 12.
Length of minor axis = 2b = 2 × 4 = 8.
Eccentricity, e = \(\frac{c}{a}\) = \(\frac{2 \sqrt{5}}{6}\) = \(\frac{\sqrt{5}}{3}\)
Latus rectum = \(\frac{2 b^{2}}{a}\) = \(\frac{2×16}{6}\) = \(\frac{16}{3}\).

GSEB Solutions Class 11 Maths Chapter 11 Conic Sections Ex 11.3

2. Equation of ellipse is \(\frac{x^{2}}{4}\) + \(\frac{y^{2}}{25}\) = 1.
Here, b2 = 4 ⇒ b = 2.
and a2 = 25 ⇒ a = 5.
Major axis is along y-axis.
c2 = 25 – 4 = 21
∴ c = \(\sqrt{21}\).
Co-ordinates of foci are (0, ± c), i.e., (0, ± \(\sqrt{21}\)).
Vertices are (0, ± a) i.e., (0, ± 5).
Length of major axis = 2a = 2 × 5 = 10.
Length of minor axis = 2b = 2 × 2 = 4.
Eccentricity e = \(\frac{c}{a}\) = \(\frac{\sqrt{21}}{5}\)
Latus rectum = \(\frac{2 b^{2}}{a}\) = \(\frac{2×4}{5}\) = \(\frac{8}{5}\).

GSEB Solutions Class 11 Maths Chapter 11 Conic Sections Ex 11.3

3. Equation of ellipse is \(\frac{x^{2}}{16}\) + \(\frac{y^{2}}{9}\) = 1.
Here, a2 = 16 ⇒ a = 4 and b2 = 9 ⇒ b = 3.
Major axis is along x-axis.
Also, c2 = a2 – b2 = 16 – 9 = 7 ⇒ c = \(\sqrt{7}\)
Co-ordinates of foci (± c, 0), i.e., (± \(\sqrt{7}\), 0).
Vertices are (± a, 0), i.e., (± 4, 0).
Length of major axis = 2a = 2 × 4 = 8.
Length of minor axis = 2b = 2 × 3 = 6.
∴ Eccentricity e = \(\frac{c}{a}\) = \(\frac{\sqrt{7}}{4}\)
Also, Latus rectum = \(\frac{2 b^{2}}{a}\) = \(\frac{2×9}{4}\) = \(\frac{9}{2}\).

4. Equation of ellipse is \(\frac{x^{2}}{25}\) + \(\frac{y^{2}}{100}\) = 1.
Major axis is along y-axis.
a2 = 100 ⇒ a = 10, b2 = 25 ⇒ b = 5.
∴ c2 = a2 – b2 = 100 – 25 = 75
∴ c = 5\(\sqrt{3}\)
Foci are (0, ±c), i.e; (0, ± 5\(\sqrt{3}\)).
Vertices are (0, ±c), i.e; (0, ±5\(\sqrt{3}\)).
Length of major axis = 2a = 2 × 10 = 20.
Length of minor axis = 2b = 2 × 5 = 10.
e = \(\frac{c}{a}\) = \(\frac{5 \sqrt{3}}{10}\) = \(\frac{\sqrt{3}}{2}\).
Length of latus rectum = \(\frac{2 b^{2}}{a}\) = \(\frac{2×25}{10}\) = 5.

GSEB Solutions Class 11 Maths Chapter 11 Conic Sections Ex 11.3

5. \(\frac{x^{2}}{49}\) + \(\frac{y^{2}}{36}\) = 1 is the equation of ellipse.
Here, major axis is along x-axis
and a2 = 49 ⇒ a = 7, b2 = 36 ⇒ b = 6.
∴ c2 = a2 – b2 = 49 – 36 = 13.
∴ c = \(\sqrt{13}\)
Foci are (± c, 0), i.e., (\(\sqrt{13}\), 0).
Vertices are (± a, 0) i.e., (± 7, 0)
Length of major axis = 2a = 2 × 7 = 14.
Length of minor axis = 2b = 2 × 6 = 12.
Length of latus rectum = \(\frac{2 b^{2}}{a}\) = \(\frac{2×36}{7}\) = \(\frac{72}{7}\)
Eccentricity, e = \(\frac{c}{a}\) = \(\frac{\sqrt{13}}{7}\).

6. \(\frac{x^{2}}{100}\) + \(\frac{y^{2}}{400}\) = 1 is the equation of the ellipse.
Major axis is along y-axis.
a2 = 400. ⇒ a = 20, b2 = 100 ⇒ b = 10.
∴ c2 = a2 – b2 = 400 – 100 = 300
∴ c = 10\(\sqrt{3}\).
Vertices are (0, ± a), i.e., (0, ± 20).
∴ Foci are (0, ± c) i.e., (0, ± 10\(\sqrt{3}\)).
Length of major axis = 2a = 2 × 20 = 40.
Length of minor axis = 2b = 2 × 10 = 20.
Eccentricity = \(\frac{c}{a}\) = \(\frac{10 \sqrt{3}}{20}\) = \(\frac{\sqrt{3}}{2}\).
Length of latus rectum = \(\frac{2 b^{2}}{a}\) = \(\frac{2×100}{20}\) = 10.

GSEB Solutions Class 11 Maths Chapter 11 Conic Sections Ex 11.3

7. 36x2 + 4y2 = 144 is the equation of the ellipse.
Dividing by 144, we get
\(\frac{x^{2}}{4}\) + \(\frac{y^{2}}{36}\) = 1.
Major axis is along y-axis, a2 = 36 or a = 6, b2 = 4 ⇒ b = 2.
∴ c2 = a2 – b2 = 36 – 4 = 32
∴ c = 4\(\sqrt{2}\).
Foci are (0, ± c), i.e., (0, ± 4\(\sqrt{2}\)).
Vertices are (0, ± a), i.e., (0 ± 6).
Length of major axis = 2a = 2 × 6 = 12.
Length of minor axis = 2b = 2 × 2 = 4.
Eccentricity e = \(\frac{c}{a}\) = \(\frac{4 \sqrt{2}}{6}\) = \(\frac{2 \sqrt{2}}{3}\).
Length of latus rectum = \(\frac{2 b^{2}}{a}\) = \(\frac{2×4}{6}\) = \(\frac{4}{3}\).

GSEB Solutions Class 11 Maths Chapter 11 Conic Sections Ex 11.3

8. The equation of the ellipse is 16x2 + y2 = 16.
Dividing by 16, we get
\(\frac{x^{2}}{1}\) + \(\frac{y^{2}}{16}\) = 1.
Major axis is along y-axis.
a2 = 16 ⇒ a = 4, b2 = 1 ⇒ b = 1.
and so c2 = a2 – b2 = 16 – 1 = 15.
∴ c = \(\sqrt{15}\).
Vertices are (0, ± a), i.e., (0, ± 4).
Length of major axis = 2a = 2 × 4 = 8.
Length of minor axis = 2b = 2 × 1 = 2.
Eccentricity e = \(\frac{c}{a}\) = \(\frac{\sqrt{15}}{4}\).
Length of latus rectum = \(\frac{2 b^{2}}{a}\) = \(\frac{2×1}{4}\) = \(\frac{1}{2}\).

9. Equation of ellipse is 4x2 + 9y2 = 36.
or \(\frac{x^{2}}{9}\) + \(\frac{y^{2}}{4}\) = 1
Major axis is along x-axis.
a2 = 9 ⇒ a = 3, b2 = 4 ⇒ b = 2.
∴ c2 = a2 – b2 = 9 – 4 = 5 ⇒ c = \(\sqrt{5}\).
Foci are (± c, 0) i.e., (± \(\sqrt{5}\), 0).
Vertices are (± a, 0), i.e., (± 3, 0).
Length of major axis = 2a = 2 × 3 = 6.
Length of minor axis = 2b = 2 × 2 = 4.
Eccentricity e = \(\frac{c}{a}\) = \(\frac{\sqrt{5}}{3}\).
Length of latus rectum = \(\frac{2 b^{2}}{a}\) = \(\frac{2×4}{3}\) = \(\frac{8}{3}\).

GSEB Solutions Class 11 Maths Chapter 11 Conic Sections Ex 11.3

In each of the following questions 10 to 20, find the equation for the ellipse that satisfies the given conditions:
10. Vertices (± 5, 0); Foci (± 4, 0)
11. Vertices (0, ± 13); Foci (0, ± 5)
12. Vertices (± 6, 0); Foci (± 4, 0)
13. Ends of major axis (± 3, 0) and ends of minor axis (0, ± 2).
14. Ends of major axis (0, ± \(\sqrt{5}\)) and ends of minor axis (± 1, 0).
15. Length of major axis 26, Foci (± 5, 0).
16. Length of minor axis = 16, Foci (0, ± 6)
17. Foci (± 3, 0); a = 4.
18. b = 3, c = 4, centre at the origin, foci on x-axis.
19. Centre at (0, 0), major axis on the y-axis and passes through the points (3, 2) and (1, 6).
20. Major axis on x-axis and passes through (4, 3) and (6, 2).
Solutions questions 10-20:
10. Vertices (± 5, 0), Foci (± 4, 0).
⇒ (± a, 0) = (± 5, 0) and (± ae, 0) = (± 4, 0).
∴ a = 5 and ae = 4.
⇒ e = \(\frac{4}{a}\) = \(\frac{4}{5}\).
Also, b2 = a2(1 – e2) [Given]
∴ b2 = 25(1 – \(\frac{16}{25}\)) – 16 = 9.
⇒ b = 3.
∴ The equation of ellipse \(\frac{x^{2}}{a^{2}}\) + \(\frac{y^{2}}{b^{2}}\) = 1 becomes
\(\frac{x^{2}}{25}\) + \(\frac{y^{2}}{9}\) = 1 ⇒ 9x2 + 25y2 = 225.
which is the equation of the required ellipse,

11. Foci (0, ± 5), vertices (0, ± 13)
(0, ± ae) = (0, ± 5) and (0, ± a) = (0, ± 13)
⇒ ae = 5 and a = 13 ∴ e = \(\frac{ae}{a}\) = \(\frac{5}{13}\).
b2 = a2 – a2e2 = 132 – 52 = 169 – 25 = 144.
∴ b = 12.
∴ Equation of the required ellipse = \(\frac{x^{2}}{144}\) + \(\frac{y^{2}}{169}\) = 1.

GSEB Solutions Class 11 Maths Chapter 11 Conic Sections Ex 11.3

12. Vertices and foci of the ellipse are (± 6, 0) and (± 4, 0) respectively.
Major axis is the x-axis.
Vertices are (± 6, 0). Foci are (± 4, 0) ⇒ c = 4 ⇒ a = 6
Now c2 = a2 – b2 or b2 = a2 – c2 = 36 – 16 = 20.
Equation of the ellipse
\(\frac{x^{2}}{36}\) + \(\frac{y^{2}}{20}\) = 1.

13. Ends of major axis are (± 3, 0).
⇒ a = 3 and major axis is x-axis.
Ends of minor axis are (0, ± 2).
⇒ b = 2.
∴ Equation of the ellipse is
\(\frac{x^{2}}{9}\) + \(\frac{y^{2}}{4}\) = 1.

14. Ends of major axis (0, ± \(\sqrt{5}\)).
Major axis is the y-axis and a2 = \(\sqrt{5}\).
Ends of minor axis are (± 1, 0)
∴ b = 1.
Equation of ellipse is
\(\frac{x^{2}}{1}\) + \(\frac{y^{2}}{5}\) = 1.

GSEB Solutions Class 11 Maths Chapter 11 Conic Sections Ex 11.3

15. Length of major axis = 2a = 26.
∴ a = 13
Foci are (± 5, 0), c = 5, ⇒ b2 = a2 – c2.
= 169 – 25
= 144.
Major axis is x-axis.
∴ Equation of ellipse is
\(\frac{x^{2}}{64}\) + \(\frac{y^{2}}{100}\) = 1.

16. Length of minor axis = 2b = 16 ⇒ b = 8.
Foci are (0, ± 6) ⇒ c = 6
∴ a2 = b2 + c2
= 64 + 36
= 100.
Major axis is y-axis.
∴ Equation of ellipse
\(\frac{x^{2}}{64}\) + \(\frac{y^{2}}{100}\) = 1.

17. Foci are (± 3, 0) ⇒ c = 3.
Also a = 4
∴ b2 = a2 – c2 = 16 – 9 = 7.
Major axis is x-axis and focus lies on it.
Equation of ellipse is
\(\frac{x^{2}}{16}\) + \(\frac{y^{2}}{7}\) = 1.

GSEB Solutions Class 11 Maths Chapter 11 Conic Sections Ex 11.3

18. b = 3, c = 4 ⇒ a2 = b2 + c2
= 9 + 16 = 25.
Foci are on x-axis.
∴ Major axis is x-axis.
∴ Equation of ellipse is
\(\frac{x^{2}}{25}\) + \(\frac{y^{2}}{9}\) = 1.

19. Major axis is y-axis.
Let the ellipse be \(\frac{x^{2}}{b^{2}}\) + \(\frac{y^{2}}{a^{2}}\) = 1.
(3, 2) and (1, 6) lies on it.
GSEB Solutions Class 11 Maths Chapter 11 Conic Sections Ex 11.3 img 1
Subtracting we get,
GSEB Solutions Class 11 Maths Chapter 11 Conic Sections Ex 11.3 img 2
or 8a2 = 32b2.
∴ a2 = 4b2.
Putting this value in (1), we get
\(\frac{9}{b^{2}}\) + \(\frac{4}{4b^{2}}\) = 1 ⇒ \(\frac{10}{b^{2}}\) = 1
∴ b2 = 10.
Now, a2 = 4b2 = 4 × 10 = 40.
∴ Equation of the ellipse is
\(\frac{x^{2}}{10}\) + \(\frac{y^{2}}{40}\) = 1.

GSEB Solutions Class 11 Maths Chapter 11 Conic Sections Ex 11.3

20. Major axis is x-axis.
Let the equation of the ellipse be
\(\frac{x^{2}}{a^{2}}\) + \(\frac{y^{2}}{b^{2}}\) = 1.
(4, 3) and (6, 2) lies on it.
Therefore,
GSEB Solutions Class 11 Maths Chapter 11 Conic Sections Ex 11.3 img 3
Subtracting (2) from (1), we get
\(\frac{- 20}{a^{2}}\) + \(\frac{5}{b^{2}}\) = 0
or 5a2 = 20b2 or a2 = 4b2.
Putting a2 = 5b2 in (1), we get
\(\frac{16}{4b^{2}}\) + \(\frac{9}{b^{2}}\) = 1 ⇒ b2 = 13
and a2 = 4b2 = 4 × 13 = 52.
∴ Equation of the ellipse is
\(\frac{x^{2}}{52}\) + \(\frac{y^{2}}{13}\) = 1.

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