If the normal to the curve y^2 = 5x - 1 at th | vedclass

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Question

(A) Given the curve equation $y^2 = 5x - 1$. Differentiating both sides with respect to $x$,we get $2y \frac{dy}{dx} = 5$,which implies $\frac{dy}{dx}

Vedclass · Accepted Answer

$4, -14$

Vedclass · Answer

(A) Given the curve equation $y^2 = 5x - 1$. Differentiating both sides with respect to $x$,we get $2y \frac{dy}{dx} = 5$,which implies $\frac{dy}{dx} = \frac{5}{2y}$. At the point $(1, -2)$,the slope of the tangent is $m_t = \frac{5}{2(-2)} = -\frac{5}{4}$. The slope of the normal $m_n$ is the negative reciprocal of the tangent slope,so $m_n = -\frac{1}{m_t} = -\frac{1}{-5/4} = \frac{4}{5}$. The equation of the normal at $(1, -2)$ is given by $y - y_1 = m_n(x - x_1)$. Substituting the values,$y - (-2) = \frac{4}{5}(x - 1)$. $5(y + 2) = 4(x - 1) \implies 5y + 10 = 4x - 4$. Rearranging the terms,$4x - 5y - 14 = 0$. Comparing this with $ax - 5y + b = 0$,we find $a = 4$ and $b = -14$.

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:

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