NCERT Class XI Mathematics - Conic Sections - Solutions

© examsnet.com
Question : 34
Total: 71
36x2+4y2 = 144
Solution:  
Given equation of ellipse is 36x2+4y2 = 144
i.e.,
36x2
144
+
4y2
144
= 1 ⇒
x2
4
+
y2
36
= 1
Clearly, 36 > 4
The equation of ellipse in standard form is
y2
a2
+
x2
b2
= 1
∴ a2 = 36 ⇒ a = 6 and b2 = 4 ⇒ b = 2
We know that c = √a2−b2 ⇒ c = √36−4 = √32 = 4√2
∴ Coordinates of foci are (0, ± c)
i.e. (0, ± 4 √2)
Coordinates of 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) =
c
a
=
4√2
6
=
2√2
3

Length of latus rectum =
2b2
a
=
2×4
6
=
4
3

© examsnet.com
Go to Question: