Question

The line x + ky + k² = 0, where k is a constant, is a tangent to the curve y² = 4x at the point P. Find, in terms of k, the coordinates of P.​

Answer 1

Answer:

The line x + ky + k² = 0, where k is a constant, is a tangent to the curve y² = 4x at the point P. Find, in terms of k, the coordinates of P.

Answer 2

Let x-ordinates of point P be a. Therefore, y coordinate = 2√a

Equation of tangent to the curve y² = 4x at point P(a,b) :-

$\rm \longrightarrow y \: y_1 = 4 \bigg( \dfrac{x + x_1}{2} \bigg ) \\ \\ \\ \rm \: put \: x_1 =a \: and \: y_1=2 \sqrt{a} \\ \\ \\ \rm \longrightarrow 2 \sqrt{a} \:y = 4 \bigg( \dfrac{x +a}{2} \bigg )\\ \\ \\ \rm \longrightarrow 2 \sqrt{a} \: y = 2 \bigg( \dfrac{x +a}{1} \bigg )\\ \\ \\ \rm \longrightarrow2 \sqrt{a} y = 2 (x +a)\\ \\ \\ \rm \longrightarrow \sqrt{a} \: y = x +a\\ \\ \\ \rm \longrightarrow \: x+a-\sqrt{a}\:y= 0......(1)$

Therefore, equation (1) is equation if tangent to the curve at point P(a,b).

But in the question it is given that The line x + ky + k² = 0 is a tangent to the curve y² = 4x at the point P.

Let , x + ky + k² = 0 be equation (2).

By comparing (1) and (2) , we get :-

➝ a = k²

➝ y₁ = 2√a

➝ y₁ = 2√k²

➝ y₁ = ±2k

Therefore, coordinates of point P = ( k², ±2k)