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)  ?

Answer 1

\[y= x^4 - b x^3 + 13 x^2 - 10x + 5\]

\[\text { Differentiating both sides w.r.t.x,} \]

\[\frac{dy}{dx} = 4 x^3 - 3b x^2 + 26x - 10\]

\[\text { Slope of tangent},m= \left( \frac{dy}{dx} \right)_\left( 0, 5 \right) =-10\]

\[\text { Given } \left( x_1 , y_1 \right) = \left( 0, 5 \right)\]

\[\text { Equation of tangent is },\]

\[y - y_1 = m \left( x - x_1 \right)\]

\[ \Rightarrow y - 5 = - 10\left( x - 0 \right)\]

\[ \Rightarrow y - 5 = - 10x\]

\[ \Rightarrow y + 10x - 5 = 0\]

\[\text { Equation of normal is},\]

\[y - y_1 = \frac{- 1}{m} \left( x - x_1 \right)\]

\[ \Rightarrow y - 5 = \frac{1}{10} \left( x - 0 \right)\]

\[ \Rightarrow 10y - 50 = x\]

\[ \Rightarrow x - 10y + 50 = 0\]