Chapter 11 Conic Sections Ex 11.2 Solutions
Question - 11 : - In each of the Exercises 7 to 12, find the equation of the parabola that satisfies the given conditions:Vertex (0, 0)passing through (2, 3) and axis is along x-axis.
Answer - 11 : -
We know that the vertex is (0, 0) and the axis of the parabolais the x-axis
The equation of the parabola is either of the from y2 = 4ax or y2 =-4ax.
Given that the parabola passes through point (2, 3), which liesin the first quadrant.
So, the equation of the parabola is of the form y2 = 4ax, while point (2, 3) mustsatisfy the equation y2 = 4ax.
Then,
32 = 4a(2)
32 = 8a
9 = 8a
a = 9/8
Thus, the equation of the parabola is
y2 = 4 (9/8)x
= 9x/2
2y2 = 9x
∴ Theequation of the parabola is 2y2 =9x
Question - 12 : - In each of the Exercises 7 to 12, find the equation of the parabola that satisfies the given conditions:Vertex (0, 0), passing through (5, 2) and symmetric with respect to y-axis.
Answer - 12 : -
We know that the vertex is (0, 0) and the parabola is symmetricabout the y-axis.
The equation of the parabola is either of the from x2 = 4ay or x2 =-4ay.
Given that the parabola passes through point (5, 2), which liesin the first quadrant.
So, the equation of the parabola is of the form x2 = 4ay, while point (5, 2) mustsatisfy the equation x2 = 4ay.
Then,
52 = 4a(2)
25 = 8a
a = 25/8
Thus, the equation of the parabola is
x2 = 4 (25/8)y
x2 = 25y/2
2x2 = 25y
∴ Theequation of the parabola is 2x2 =25y