**Equation of Parabola**

Equation of Parabola when vertex is (h, k) and Axis is Parallel to Y-axis.

The equation of the parabola with vertex (h, k) is (x – h)² = 4a (y – k)

Its focus is at (h, a + k) and the length of the latus rectum is 4a.

The equation of the directrix is y = k – a

(or) y + a – k = 0

y + a – k = 0

**Parabolic Curve: **The equations y = Ax² + Bx + C and x = Ay² + By + C (A ≠ 0) represent parabolas and are called parabolic curves.

Now,

y = Ax² + Bx + C

\(=A\left( {{x}^{2}}+\frac{B}{A}x+\frac{C}{A}\right)=A\left\{ {{\left( x+\frac{B}{2A}\right)}^{2}}-\frac{{{B}^{2}}}{4{{A}^{2}}}+\frac{C}{A} \right\}\).

\(=A\left\{ {{\left( x+\frac{B}{2A}\right)}^{2}}-\frac{\left( {{B}^{2}}-4AC \right)}{4{{A}^{2}}} \right\}\).

Or \({{\left( x+\frac{B}{2A}\right)}^{2}}=\frac{1}{A}\left( y+\frac{{{B}^{2}}-4AC}{4A} \right)\).

Comparing it with (x – h)² = 4a (y – k), it represents a parabola with vertex (h, k) = (-B/ 2A, -[B² – 4AQ]/ 4A) axis parallel to the y-axis, latus return = 1, and the curve opening upwards and downwards depending upon the sign of A (for A > 0,the curve opens upward; for A < 0 curve opens downward).

Similarly, x = Ay² + By + C can be simplified to

\({{\left( y+\frac{B}{2A}\right)}^{2}}=\frac{1}{A}\left( x+\frac{{{B}^{2}}-4AC}{4A} \right)\).

Comparing it with (y – k)² = 4a (x – h), It represents a parabola with vertex at (h, k) =(-[B² – 4AC]/ 4A , – B/ 2A), axis parallel to the x-axis, latus rectum = 1/|A|, and the curve opening left and right according to A < 0 and A > 0, respectively.

**Note: **The parametric form of the parabola (y – k) ² = 4a (x – h) is x = h + at² and y = k + 2at

**Example: ** Find the value of p such that the vertex of y = x² + 2px + 13 is 4 units above the x-axis.

**Solution: **The given parabola is y = x² + 2px + 13

y = (x + p) ² + 13 – p²

y – (13 – p²) = (x + p) ²

So, the vertex is (-p, 13 – p²).

Since the parabola is at a distance of 4 units above the x-axis,

13 – p² = 4

p² = 9

p = ± 3.