Question

If the normal to the curve $y^2 = 5x - 1$, at the point (1,- 2) is of the form ax - 5y + b = 0, then a and b are:

• A. 4, - 14
• B. 44665
• C. - 4, 14
• D. - 4, - 14

Answer 1

Answer: (A)

4, - 14

Answer 2

Equation of curve : $y^2 = 5 x - 1$ Differentiating w.r. to x, we get $\frac{dy}{dx} = \frac{5}{2y}$ $\Rightarrow \, \left( \frac{dy}{dx} \right)_{(1, -2)} = - \frac{5}{4}$ $\Rightarrow$ slope of normal $= \frac{4}{5}$ Now, equation of normal to the curve at (1,- 2) is given by $y + 2 = \frac{4}{5} (x - 1)$ or 4x - 5y - 14 = 0 ....(1) Comparing the coefficients of like terms of this equation and the given equation to the normal, we obtain. a = 4, b = - 14