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Q.
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:
Application of Derivatives
Solution:
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