Find the equation of the tangent and the normal to the following curves at the indicated points.
$y = x^{4} - b x^{3} + 13 x^{2} - 10 x + 5$ at $(0, 5)$ .

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Solution

Tangent $= y + 10 x - 5 = 0$
Normal $= x - 10 y + 50 = 0$ .
The equation of the curve is $y = x^{4} - b x^{3} + 13 x^{2} - 10 x + 5$ .
$∴ \frac{d y}{d x} = 4 x^{3} - 3 b x^{2} + 26 x - 10$
$\Rightarrow {(\frac{d y}{d x})}_{(0, 5)} = - 10$
The equation of tangent at $(0, 5)$ is
$y - 5 = {(\frac{d y}{d x})}_{(0, 5)} (x - 0)$
$\Rightarrow y - 5 = - 10 (x - 0)$
$\Rightarrow 10 x + y - 5 = 0$
The equation of the normal at $(0, 5)$ is
$y - 5 = - {\frac{1}{(\frac{d y}{d x})}}_{(0, 5)} (x - 0)$
$\Rightarrow y - 5 = \frac{1}{10} (x - 0)$
$\Rightarrow x - 10 y + 50 = 0$ .

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