# 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.

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