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We will learn how to find the position of a point with respect to a parabola.
The position of a point (x1, y1) with respect to a parabola y2 = 4ax (i.e. the point lies outside, on or within the parabola) according as y12 - 4ax1 >, =, or < 0.
Let P(x1, y1) 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 y2 = 4ax at Q and let the coordinates of Q be (x1, y2). Now, the point Q (x1, y2) lies on the parabola y2 = 4ax. Hence we get
y22 = 4ax1
Therefore, the point
(i) P lies outside the parabola y2 = 4ax if PN > QN
i.e., PN2 > QN2
⇒ y12 > y22
⇒ y12 > 4ax1, [Since, 4ax1 = y22].
(ii) P lies on the parabola y2 = 4ax if PN = QN
i.e., PN2 = QN2
⇒ y12 = y22
⇒ y12 = 4ax1, [Since, 4ax1 = y22].
(iii) P lies outside the parabola y2 = 4ax if PN < QN
i.e., PN2 < QN2
⇒ y12 < y22
⇒ y12 < 4ax1, [Since, 4ax1 = y22].
Therefore, the point P (x1, y1) lies outside, on or within the parabola y2 = 4ax according as
y12 - 4ax1 >,= or < 0.
Notes:
(i) The point P(x1, y1) lies outside, on or within the parabola y2 = -4ax according as y12 + 4ax1 >, = or <0.
(ii) The point P(x1, y1) lies outside, on or within the parabola x2 = 4ay according as x12 - 4ay1 >, = or <0.
(ii) The point P(x1, y1) lies outside, on or within the parabola x2 = -4ay according as x12 + 4ay1 >, = or <0.
Solved examples to find the position of the point P (x1, y1) with respect to the parabola y2 = 4ax:
1. Does the point (-1, -5) lies outside, on or within the parabola y2 = 8x?
Solution:
We know that the point (x1, y1) lies outside, on or within the parabola y2 = 4ax according as y12 - 4ax1 is positive, zero or negative.
Now, the equation of the given parabola is y2 = 8x ⇒ y2 - 8x= 0
Here x1 = -1 and y1 = -5
Now, y12 - 8x1 = (-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 y2 = 12x but the point (- 2, - 3) lies within it."
Solution:
We know that the point (x1, y1) lies outside, on or within the parabola y2 = 4ax according as y12 - 4ax1 is positive, zero or negative.
Now, the equation of the given parabola is y2 = 12x or, y2 - 12x = 0
For then point (2, 3):
Here x1 = 2 and y1 = 3
Now, y12 - 12x1 = 32 – 12 ∙ 2 = 9 - 24 = -15 < 0
Hence, the point (2, 3) lies within the parabola y2 = 12x.
For then point (-2, -3):
Here x1 = -2 and y1 = -3
Now, y12 - 12x1 = (-3)2 – 12 ∙ (-2) = 9 + 24 = 33 > 0
Hence, the point (-2, -3) lies outside the parabola y2 = 12x.
Therefore, the given statement is not valid.
● The Parabola
11 and 12 Grade Math
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