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. 4, 14
• C. - 4, 14
• D. - 4, - 14

Answer 1

Answer: (A)

Equation of curve : $y^2=5x-1$
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