We will learn how to find the position of a point with respect to a parabola.
The position of a point (x\(_{1}\), y\(_{1}\)) with respect to a parabola y\(^{2}\) = 4ax (i.e. the point lies outside, on or within the parabola) according as y\(_{1}\)\(^{2}\) - 4ax\(_{1}\) >, =, or < 0.
Let P(x\(_{1}\), y\(_{1}\)) be a point on the plane. From P draw PN perpendicular to the x-axis i.e., AX and N being the foot of the perpendicular.
PN intersect the parabola y\(^{2}\) = 4ax at Q and let the coordinates of Q be (x\(_{1}\), y\(_{2}\)). Now, the point Q (x\(_{1}\), y\(_{2}\)) lies on the parabola y\(^{2}\) = 4ax. Hence we get
y\(_{2}\)\(^{2}\) = 4ax\(_{1}\)
Therefore, the point
(i) P lies outside the parabola y\(^{2}\) = 4ax if PN > QN
i.e., PN\(^{2}\) > QN\(^{2}\)
⇒ y\(_{1}\)\(^{2}\) > y\(_{2}\)\(^{2}\)
⇒ y\(_{1}\)\(^{2}\) > 4ax\(_{1}\), [Since, 4ax\(_{1}\) = y\(_{2}\)\(^{2}\)].
(ii) P lies on the parabola y\(^{2}\) = 4ax if PN = QN
i.e., PN\(^{2}\) = QN\(^{2}\)
⇒ y\(_{1}\)\(^{2}\) = y\(_{2}\)\(^{2}\)
⇒ y\(_{1}\)\(^{2}\) = 4ax\(_{1}\), [Since, 4ax\(_{1}\) = y\(_{2}\)\(^{2}\)].
(iii) P lies outside the parabola y\(^{2}\) = 4ax if PN < QN
i.e., PN\(^{2}\) < QN\(^{2}\)
⇒ y\(_{1}\)\(^{2}\) < y\(_{2}\)\(^{2}\)
⇒ y\(_{1}\)\(^{2}\) < 4ax\(_{1}\), [Since, 4ax\(_{1}\) = y\(_{2}\)\(^{2}\)].
Therefore, the point P (x\(_{1}\), y\(_{1}\)) lies outside, on or within the parabola y\(^{2}\) = 4ax according as
y\(_{1}\)\(^{2}\) - 4ax\(_{1}\) >,= or < 0.
Notes:
(i) The point P(x\(_{1}\), y\(_{1}\)) lies outside, on or within the parabola y\(^{2}\) = -4ax according as y\(_{1}\)\(^{2}\) + 4ax\(_{1}\) >, = or <0.
(ii) The point P(x\(_{1}\), y\(_{1}\)) lies outside, on or within the parabola x\(^{2}\) = 4ay according as x\(_{1}\)\(^{2}\) - 4ay\(_{1}\) >, = or <0.
(ii) The point P(x\(_{1}\), y\(_{1}\)) lies outside, on or within the parabola x\(^{2}\) = -4ay according as x\(_{1}\)\(^{2}\) + 4ay\(_{1}\) >, = or <0.
Solved examples to find the position of the point P (x\(_{1}\), y\(_{1}\)) with respect to the parabola y\(^{2}\) = 4ax:
1. Does the point (-1, -5) lies outside, on or within the parabola y\(^{2}\) = 8x?
Solution:
We know that the point (x\(_{1}\), y\(_{1}\)) lies outside, on or within the parabola y\(^{2}\) = 4ax according as y\(_{1}\)\(^{2}\) - 4ax\(_{1}\) is positive, zero or negative.
Now, the equation of the given parabola is y\(^{2}\) = 8x ⇒ y\(^{2}\) - 8x= 0
Here x\(_{1}\) = -1 and y\(_{1}\) = -5
Now, y\(_{1}\)\(^{2}\) - 8x\(_{1}\) = (-5)\(^{2}\) - 8 ∙ (-1) = 25 + 8 = 33 > 0
Therefore, the given point lies outside the given parabola.
2. Examine with reasons the validity of the following statement:
"The point (2, 3) lies outside the parabola y\(^{2}\) = 12x but the point (- 2, - 3) lies within it."
Solution:
We know that the point (x\(_{1}\), y\(_{1}\)) lies outside, on or within the parabola y\(^{2}\) = 4ax according as y\(_{1}\)\(^{2}\) - 4ax\(_{1}\) is positive, zero or negative.
Now, the equation of the given parabola is y\(^{2}\) = 12x or, y\(^{2}\) - 12x = 0
For then point (2, 3):
Here x\(_{1}\) = 2 and y\(_{1}\) = 3
Now, y\(_{1}\)\(^{2}\) - 12x\(_{1}\) = 3\(^{2}\) – 12 ∙ 2 = 9 - 24 = -15 < 0
Hence, the point (2, 3) lies within the parabola y\(^{2}\) = 12x.
For then point (-2, -3):
Here x\(_{1}\) = -2 and y\(_{1}\) = -3
Now, y\(_{1}\)\(^{2}\) - 12x\(_{1}\) = (-3)\(^{2}\) – 12 ∙ (-2) = 9 + 24 = 33 > 0
Hence, the point (-2, -3) lies outside the parabola y\(^{2}\) = 12x.
Therefore, the given statement is not valid.
● The Parabola
11 and 12 Grade Math
From Position of a Point with respect to a Parabola to HOME PAGE
Didn't find what you were looking for? Or want to know more information about Math Only Math. Use this Google Search to find what you need.
Aug 14, 24 09:52 AM
Aug 14, 24 02:39 AM
Aug 13, 24 01:27 AM
Aug 12, 24 03:20 PM
Aug 12, 24 02:23 AM
New! Comments
Have your say about what you just read! Leave me a comment in the box below. Ask a Question or Answer a Question.