Page 11 - LESSON NOTES

5. Find the equation of the parabola with focus (2,0) and directrix x = – 2.
Soln: Since the focus (2,0) lies on the x-axis, the x-axis itself is the axis of the

parabola.
2
2
Hence the equation of the parabola is of the form either y = 4ax or y = – 4ax.
2
Since the directrix is x = – 2 and the focus is (2,0), the parabola is to be of the form y
= 4ax with a = 2.
Hence the required equation is y2 = 4(2)x = 8x
6. Find the equation of the parabola which is symmetric about the y-axis, and passes
through the point (2,–3)
Soln: Since the parabola is symmetric about y-axis and has its vertex at the origin, the
2
2
equation is of the form x = 4ay or x = – 4ay, where the sign depends on whether
the parabola opens upwards or downwards.
But the parabola passes through (2,–3) which lies in the fourth quadrant, it must

open downwards.
2
Thus the equation is of the form x = – 4ay.
Since the parabola passes through ( 2,–3), we have
1
2
2 = – 4a (–3), i.e., a =
3
1
2
2
Therefore, the equation of the parabola is x = −4 ( ) y. i.e 3x = – 4y
3
7. Find the coordinates of the foci, the vertices, the lengths of major and minor axes
2
2
and the eccentricity of the ellipse 9 x + 4y = 36.
Soln: The given equation of the ellipse can be written in standard form as
x 2 y 2
+ = 36
4 9
y 2 x 2
Since the denominator of is larger than the denominator of , the major axis is
9 4
along the y-axis.
x 2 y 2
Comparing the given equation with the standard equation + = 1
b 2 a 2
we have b = 2 and a = 3
2
2
Also c = √a − b = √5
c √5
e = =
a 3
Hence the foci are (0, √5 ) and (0, – √5 ), vertices are (0,3) and (0, –3), length of the
major axis is 6 units, the length of the minor axis is 4 units and the eccentricity of the
√5
ellipse is
3