The slope of the tangent at any point (x, y) | vedclass

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(C) The slope of the tangent at any point $(x, y)$ is given by $\frac{dy}{dx} = xy$. At the point $(\sqrt{2}, e)$,the slope of the tangent is $m = \sq

Vedclass · Accepted Answer

$\sqrt{2}e$

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(C) The slope of the tangent at any point $(x, y)$ is given by $\frac{dy}{dx} = xy$. At the point $(\sqrt{2}, e)$,the slope of the tangent is $m = \sqrt{2}e$. The slope of the normal at this point is $m' = -\frac{1}{m} = -\frac{1}{\sqrt{2}e}$. The equation of the normal at $(\sqrt{2}, e)$ is given by $(y - e) = m'(x - \sqrt{2})$. Substituting $m'$,we get $y - e = -\frac{1}{\sqrt{2}e}(x - \sqrt{2})$. Multiplying by $\sqrt{2}e$,we get $\sqrt{2}ey - \sqrt{2}e^2 = -x + \sqrt{2}$. Rearranging the terms,$x + \sqrt{2}ey = \sqrt{2} + \sqrt{2}e^2$. Dividing by $\sqrt{2}(1 + e^2)$,we get $\frac{x}{\sqrt{2}(1 + e^2)} + \frac{\sqrt{2}ey}{\sqrt{2}(1 + e^2)} = 1$. Comparing this with $ax + by = 1$,we have $a = \frac{1}{\sqrt{2}(1 + e^2)}$ and $b = \frac{\sqrt{2}e}{\sqrt{2}(1 + e^2)} = \frac{e}{1 + e^2}$. Therefore,$\frac{b}{a} = \frac{e}{1 + e^2} 	imes \sqrt{2}(1 + e^2) = \sqrt{2}e$.

The slope of the tangent at any point $(x, y)$ on the curve is equal to the product of the coordinates of that point. If the equation of the normal to the curve at the point $(\sqrt{2}, e)$ is $ax + by = 1$,then $\frac{b}{a} =$

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