**Equation of Pair of Tangents from Point – Parabola**

Let T (h, k) be nay point on the pair of tangents PQ or PR draw P (x₁, y₁) to the parabola y² = 4ax.

The equation of PT is y – y₁ = (k – y₁)/ (h – x₁) (x – x₁)

(or)

y = ((k – y₁)/ (h – x₁)) x + ((hy₁ – kx₁)/ (h – x₁))

Which is tangent to the parabola y² = 4ax.

∴ c = a/m

cm = a

\(\left( \frac{h{{y}_{1}}-k{{x}_{1}}}{h-{{x}_{1}}} \right)\left( \frac{k-{{y}_{1}}}{h-{{x}_{1}}} \right)=a\).

(k – y₁) (hy₁ – kx₁) = a(h – x₁)²

Therefore, the locus of (h, k) is

(y – y₁) (xy₁ – x₁y) = (x-x₁)²

(y² – 4ax) (y₁² – 4ax₁) = {(yy₁ – 2a (x + x₁)}²

SS₁ = T²

Where,

S = y² = 4ax

S₁ = y₁² = 4ax₁

and T = yy₁ – 2a (x + x₁)

**Example:** find the equation of the tangent to the parabola y² = 8x having slope 2 and also find the point of contact.

**Solution: **y² = 8x … (1)

y² = 4ax … (2)

compare equation (1) and (2) we will get

a = 2

The equation of the tangent to y² = 4ax having slope m is

Y = mx + (a/m)

Given that m = 2

Y = 2x + (2/2)

Y = 2x + 1

Point of contact is \(\left( \frac{a}{{{m}^{2}}},\frac{2a}{m} \right)=\left( \frac{2}{{{2}^{2}}},\frac{2.2}{2} \right)=\left( \frac{1}{2},2 \right)\).