Page 12 - LESSON NOTES

8. Find the equation of the ellipse whose vertices are (±13, 0) and foci are (±5, 0).
Soln : Since the vertices are on x-axis, the equation will be of the form
x 2 y 2
+ = 1 where a is the semi-major axis.
a 2 b 2
2
2
2
Given that a = 13, c = ± 5. Therefore, from the relation = − we get
2
25 = 169 − ⇒ = 12
x 2 y 2
Hence the equation of the ellipse is + = 1
169 144
9. Find the equation of the ellipse, whose length of the major axis is 20 and foci are
(0, ± 5).
Soln : Since the foci are on y-axis, the major axis is along the y-axis. So, equation of
x 2 y 2
the ellipse is of the form + = 1 .
b 2 a 2
20
2
2
2
Given that a = semi-major axis = = 10 and the relation = − gives
2
2
2
2
2
5 = 10 − ⇒ = 75
x 2 y 2
Therefore, the equation of the ellipse is + = 1
75 100
10. Find the coordinates of the foci and the vertices, the eccentricity, the length of the
x 2 y 2
latus rectum of the hyperbola − = 1.
9 16
x 2 y 2 x 2 y 2
Soln : Comparing the equation − = 1 with the standard equation − = 1 ,
9 16 a 2 b 2
2
2
Here a = 3, b = 4 and c = √a + b = 5
Therefore, the coordinates of the foci are (± 5, 0) and that of vertices are (± 3, 0).Also,
c 5 2b 2 32
The eccentricity e = = . The latus rectum = = .
a 3 a 3
11. Find the equation of the hyperbola where foci are (0, ±12) and the length of the latus
rectum is 36.
Soln: Since foci are (0, ± 12), it follows that c = 12.
2b 2
2
Length of the latus rectum = = 36 or b = 18a
a
2
2
2
2
Therefore c = a + b gives 144 = a + 18a
2
⇒a + 18a−144 = 0
⇒a = −24,6
2
Since a cannot be negative, we take a = 6 and so b = 108.
y 2 x 2
Therefore, the equation of the required hyperbola is − = 1.
36 108