**Position of the Point with respect to the Parabola**

Let S ≡ y² = 4ax be the equation of the parabola and P(x₁, y₁), be any in the region of the parabola, then S₁ ≡ y₁² – 4ax₁.

The point (x₁, y₁) lies outside, on or within the parabola y² = 4ax according as S₁ ≡ y₁² – 4ax₁ is positive, zero or negative

If S₁ > 0, then P lies outside the parabola

If S₁ = 0, then P lies on the parabola

If S₁ < 0, then P lies inside the parabola

**Examples:**

**1)** The position of point (1, 4) with respect to the parabola y² + 9 – 6y =5x

**Solution: **Let S = y² + 9 – 6y – 5x = 0

At point P (1, 4)

Let S ≡ y² = 4ax be the equation of the parabola and P (x₁, y₁), be any in the region of the parabola, then S₁ ≡ y₁² – 4ax₁

S₁ = y₁² + 9 – 6y₁ – 5x₁

S₁ = (4)² + 9 – 6(4) – 5(1)

S₁ = 16 + 9 – 24 – 5

S₁ = -4 < 0

S₁ < 0, then P lies inside the parabola

Hence given point lies inside the curve

**2)** Does the point (1, 3) lies outside, on or within the parabola y² = 8x

**Solution: **The point P (x₁, y₁) lies outside, on or within the parabola y² = 4ax according as S₁ ≡ 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 = 3

Now, y₁² – 8x₁ = (3)² – 8 (1) = 9 – 8 = 1 > 0

Therefore, the given point lies outside the given parabola.