Question

Find the equation of the tangent to y²=4ax at(x₁,y₁).

Answer 1

equation of the curve is given, y² = 4ax.

we have to find equation of tangent of the curve at (x₁,y₁).

first, differentiate curve y² = 4ax with respect to x,

⇒2y dy/dx = 4a

⇒dy/dx = 2a/y

so, equation of slope of tangent of the curve is m = $\frac{dy}{dx}|_{(x_1,y_1)}$ = 2a/y₁

now, equation of tangent is ...

(y - y₁) = m(x - x₁)

⇒(y - y₁) = 2a/y₁(x - x₁)

⇒yy₁ - y₁² = 2ax - 2ax₁

⇒yy₁ -2ax + 2ax₁ - y₁² = 0 ....(1)

putting (x₁,y₁) on the curve. we get, y₁² = 4ax₁......(2)

putting equation (2) in equation (1),

we get, yy₁ - 2ax + 2ax₁ - 4ax₁ = 0

⇒yy₁ - 2ax - 2ax₁ = 0

⇒yy₁ = 2a(x + x₁), This is the equation of tangent of the curve, y² = 4ax at (x₁,y₁).