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Question
Find the equation of the tangent and the normal to the following curve at the indicated point y = x4 − bx3 + 13x2 − 10x + 5 at (0, 5) ?
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Solution
\[y= x^4 - b x^3 + 13 x^2 - 10x + 5\]
\[\text { Differentiating both sides w.r.t.x,} \]
\[\frac{dy}{dx} = 4 x^3 - 3b x^2 + 26x - 10\]
\[\text { Slope of tangent},m= \left( \frac{dy}{dx} \right)_\left( 0, 5 \right) =-10\]
\[\text { Given } \left( x_1 , y_1 \right) = \left( 0, 5 \right)\]
\[\text { Equation of tangent is },\]
\[y - y_1 = m \left( x - x_1 \right)\]
\[ \Rightarrow y - 5 = - 10\left( x - 0 \right)\]
\[ \Rightarrow y - 5 = - 10x\]
\[ \Rightarrow y + 10x - 5 = 0\]
\[\text { Equation of normal is},\]
\[y - y_1 = \frac{- 1}{m} \left( x - x_1 \right)\]
\[ \Rightarrow y - 5 = \frac{1}{10} \left( x - 0 \right)\]
\[ \Rightarrow 10y - 50 = x\]
\[ \Rightarrow x - 10y + 50 = 0\]
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