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 (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.

$Position of a Point with respect to a Parabola$ Position of a Point with respect to a Parabola

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

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