Where Mindful Learning Begins
Where Mindful Learning Begins
Free, step-by-step Class 11 Maths NCERT Solutions for Chapter 10 Ex 10.2 — all 12 questions solved, covering the Parabola: reading off the focus, axis, directrix and latus rectum from a standard equation, and finding the equation of a parabola from its focus, directrix, or a point it passes through.
There are exactly four standard forms of a parabola with vertex at the origin — y^2=\pm4ax and x^2=\pm4ay — and every question in this exercise is about matching a given equation, or a given focus and directrix, to the right one of these four. Once the correct form and the value of a are identified, the focus, directrix and latus rectum all follow directly from the standard results.
The equation has a y^2 term with a positive coefficient of x, so it is of the form y^2=4ax, with the axis along the x-axis and the parabola opening to the right.
Comparing with y^2=4ax: 4a=12 \ \Rightarrow\ a=3.
The equation has an x^2 term with a positive coefficient of y, so it is of the form x^2=4ay, with the axis along the y-axis and the parabola opening upward.
Comparing with x^2=4ay: 4a=6 \ \Rightarrow\ a=\dfrac32.
The equation has a y^2 term with a negative coefficient of x, so it is of the form y^2=-4ax, with the axis along the x-axis and the parabola opening to the left.
Comparing with y^2=-4ax: 4a=8 \ \Rightarrow\ a=2.
The equation has an x^2 term with a negative coefficient of y, so it is of the form x^2=-4ay, with the axis along the y-axis and the parabola opening downward.
Comparing with x^2=-4ay: 4a=16 \ \Rightarrow\ a=4.
The equation is of the form y^2=4ax, opening to the right.
Comparing with y^2=4ax: 4a=10 \ \Rightarrow\ a=\dfrac52.
The equation is of the form x^2=-4ay, opening downward.
Comparing with x^2=-4ay: 4a=9 \ \Rightarrow\ a=\dfrac94.
Since the focus (6,0) lies on the x-axis and the directrix is the vertical line x=-6, the axis of the parabola is the x-axis, and the vertex (midway between focus and directrix) is at the origin. Since the focus is to the right of the vertex, the parabola opens to the right, so the equation has the form y^2=4ax with a=6.
Since the focus (0,–3) lies on the y-axis and the directrix is the horizontal line y=3, the axis is the y-axis, with vertex at the origin. Since the focus is below the vertex, the parabola opens downward, so the equation has the form x^2=-4ay with a=3.
Since the vertex is (0,0) and the focus (3,0) lies on the positive x-axis, the parabola opens to the right, so the equation has the form y^2=4ax with a=3.
Since the vertex is (0,0) and the focus (–2,0) lies on the negative x-axis, the parabola opens to the left, so the equation has the form y^2=-4ax with a=2.
With vertex at the origin and axis along the x-axis, the equation is of the form y^2=4ax or y^2=-4ax. Since the parabola passes through (2,3), which has a positive x-coordinate, it must open to the right, so the equation is of the form y^2=4ax.
Substituting the point (2,3):
3^2=4a(2) \ \Rightarrow\ 9=8a \ \Rightarrow\ a=\dfrac98
So the equation is:
y^2=4\left(\dfrac98\right)x=\dfrac92x
Being symmetric about the y-axis with vertex at the origin means the equation is of the form x^2=4ay or x^2=-4ay. Since the parabola passes through (5,2), which has a positive y-coordinate, it must open upward, so the equation is of the form x^2=4ay.
Substituting the point (5,2):
5^2=4a(2) \ \Rightarrow\ 25=8a \ \Rightarrow\ a=\dfrac{25}{8}
So the equation is:
x^2=4\left(\dfrac{25}{8}\right)y=\dfrac{25}{2}y
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