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:

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

If the normal to the curve $y^2 = 5x

Notes on Application of derivatives

Concepts Used:

Application of Derivatives

Various Applications of Derivatives-

Rate of Change of Quantities:

If some other quantity ‘y’ causes some change in a quantity of surely ‘x’, in view of the fact that an equation of the form y = f(x) gets consistently pleased, i.e, ‘y’ is a function of ‘x’ then the rate of change of ‘y’ related to ‘x’ is to be given by

$\frac{\triangle y}{\triangle x}=\frac{y_2-y_1}{x_2-x_1}$

This is also known to be as the Average Rate of Change.

Increasing and Decreasing Function:

Consider y = f(x) be a differentiable function (whose derivative exists at all points in the domain) in an interval x = (a,b).

If for any two points x₁ and x₂ in the interval x such a manner that x₁ < x₂, there holds an inequality f(x₁) ≤ f(x₂); then the function f(x) is known as increasing in this interval.
Likewise, if for any two points x₁ and x₂ in the interval x such a manner that x₁ < x₂, there holds an inequality f(x₁) ≥ f(x₂); then the function f(x) is known as decreasing in this interval.
The functions are commonly known as strictly increasing or decreasing functions, given the inequalities are strict: f(x₁) < f(x₂) for strictly increasing and f(x₁) > f(x₂) for strictly decreasing.

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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:

The Correct Option is A

Solution and Explanation

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