Question

If the slope of the tangent to the curve y=x log x at a point on it is 3/2,then find the equations of tangent and normal at that point

Answer 1

$Given \:curve \: is \: y = x log x \: ---(1)$

$Let \: (a,b) \: be \: any \: point \: on \:the \\curve . \: [ i.e., b = log a * ]$

/* Differentiate with respect to ' x ' on bothsides of the equation (1) */

$\frac{dy}{dx} = x \times \frac{1}{x} + log x \times 1$

$\implies \frac{dy}{dx} = 1 + log x$

$\big( \frac{dy}{dx}_{(a,b)}\big) = 1 + log a$

$Given \: slope = \frac{3}{2}$

$\implies \frac{3}{2} = 1 + log a$

$\implies \frac{3}{2} - 1 = log a$

$\implies \frac{3-2}{2} = log a$

$\implies \frac{1}{2} = log_{e}a$

$\implies a = \sqrt{e} \: [ From \: * ]$

$\implies b = \sqrt{e} log \sqrt{e}$

$\implies b =\frac{1}{2} \sqrt{e} log e$

$\boxed { \pink { log a^{m} = m log a }}$

$\implies b = \frac{1}{2} \sqrt{e}$

$\boxed { \orange { log e = 1 }}$

$\underline { \blue { Equation \: of \: tangent \:is : }}$

$\big( y - \frac{1}{2} \sqrt{e}\big) = \frac{3}{2} ( x - \sqrt{e})$

$\implies 2y - \sqrt{e} = 3x - 3\sqrt{e}$

$\implies 3x - 2y = 2 \sqrt{e}$

$\underline { \blue { Finding \: Equation \: of \: normal \:is : }}$

$The \: slope \: of \: normal \: is \: \frac{-2}{3}$

$The \: equation \: of \: normal \: is :\\big(y - \frac{1}{2} \sqrt{e} \big) = \frac{-2}{3} ( x - \sqrt{e})$

$\implies 3y - \frac{3}{2} \sqrt{e} = -2x + 2 \sqrt{e}$

$\implies 2x +3y = \frac{3}{2} \sqrt{e} + 2 \sqrt{e}$

$\implies 2x + 3y = \frac{ 3\sqrt{e} + 4\sqrt{e}}{2}$

$\implies 4x + 6y = 7\sqrt{e}$

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