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Position of a Point with respect to a Parabola

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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