We will learn how to solve different types of problems on parabola.
1. Find the vertex, focus, directrix, axis and latusrectum of the parabola y\(^{2}\)  4x  4y = 0
Solution:
The given equation of the parabola is y\(^{2}\)  4x  4y = 0
⇒ y\(^{2}\)  4y = 4x
⇒ y\(^{2}\)  4y + 4 = 4x + 4, (Adding 4 on both sides)
⇒ (y  2)\(^{2}\) = 4(x + 1) ……………………………….. (i)
Shifting the origin to the point (1, 2) without rotating the
axes and denoting the new coordinates with respect to these axes by X and Y, we
have
x = X + (1), y = Y + 2 ……………………………….. (ii)
Using these relations equation (i), reduces to
Y\(^{2}\) = 4X……………………………….. (iii)
This is of the form Y\(^{2}\) = 4aX. On comparing, we get 4a = 4 ⇒ a = 1.
The coordinates of the vertex with respect to new axes are (X = 0, Y = 0)
So, coordinates of the vertex with respect to old axes are (1, 2), [Putting X= 0, Y = 0 in (ii)].
The coordinates of the focus with respect to new axes are (X = 1, Y = 0)
So, coordinates of the focus with respect to old axes are (0, 2), [Putting X= 1, Y = 0 in (ii)].
Equation of the directrix of the parabola with respect to new axes in X = 1
So, equation of the directrix of the parabola with respect to old asex is x = 2, [Putting X = 1, in (ii)].
Equation of the axis of the parabola with respect to new axes is Y = 0.
So, equation of axis with respect to old axes is y = 2, [Putting Y = 0, in (ii)].
The length of the latusrectum is 4 units.
2. Find the point on the parabola y\(^{2}\) = 12x at which the ordinate is double the abscissa.
Solution:
The given parabola is y\(^{2}\) = 12x.
Now, let (k, 2k) be the coordinates of the required point (k ≠ 0).
Since the point lies (k, 2k) on the parabola y\(^{2}\) = 12x,
Therefore, we get,
(2k)\(^{2}\) = 12k
⇒ 4k\(^{2}\) = 12k
⇒ k = 3 (Since, k ≠ 0, ).
Therefore, the coordinates of the required point are (3, 6).
3. Write the parametric equation of the parabola (x + 2)\(^{2}\) =  4(y + 1).
Solution:
The given equation of the parabola is (x + 2)\(^{2}\) =  4(y + 1).
Then parametric equation of the parabola (x + 2)\(^{2}\) =  4(y + 1) are
x + 2 = 2t and y + 1 = t\(^{2}\)
⇒ x = 2t – 2 and y = t\(^{2}\) – 1.
4. Find the equation of the parabola whose coordinates of vertex and focus are (2, 3) and (1, 3) respectively.
Solution:
According to the problem, the ordinates of vertex and focus are equal hence, the axis of the required parabola is parallel to xaxis. Again,
a = abscissa of focus  abscissa of vertex
⇒ a = 1  ( 2) = 1 + 2 = 3.
Therefore, the equation of the required parabola is
(y  β)\(^{2}\) = 4a (x  α)
⇒ (y  3)\(^{2}\) = 4 . 3(x + 2)
⇒ y\(^{2}\)  6y + 9 = 12x + 24
⇒ y\(^{2}\)  6y  12x = 15.
● The Parabola
11 and 12 Grade Math
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