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