Question

Find the equation of the normal to curve y^ 2 = 4x at the point (1, 2).

Answer 1

${y}^{2} = 4x \\ diffrentiating \: w.r.t \: x \\ 2y \frac{dy}{dx} = 4 \\ \frac{dy}{dx} = \frac{4}{2y} \\ \frac{dy}{dx} = \frac{2}{y } \\ put \: point \: (1,2) \\ \frac{dy}{dx} = 1 \\ slope \: of \: normal = \frac{ - 1}{ \frac{dy}{dx} } \\ slope \: of \: normal = \frac{ - 1}{1} = - 1 \\ equqtion \: will \: be - \\ slope = \frac{y - y1}{x - x1} \\ \\ - 1 = \frac{y -2 }{x - 1} \\ - x + 1 = y - 2 \\ x + y = 3$

Answer 2

Answer:

Step-by-step explanation:

Hope you understood.