Question

Find the equation of the tangent and the normal to the following curve at the indicated point y² = 4ax at \[\left( \frac{a}{m^2}, \frac{2a}{m} \right)\] ?

Answer 1

\[y^2 =4ax\]

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

\[2y \frac{dy}{dx} = 4a\]

\[ \Rightarrow \frac{dy}{dx} = \frac{2a}{y}\]

\[\text { Given } \left( x_1 , y_1 \right) = \left( \frac{a}{m^2}, \frac{2a}{m} \right)\]

\[\text { Slope of tangent }= \left( \frac{dy}{dx} \right)_\left( \frac{a}{m^2}, \frac{2a}{m} \right) =\frac{2a}{\left( \frac{2a}{m} \right)}=m\]

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

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

\[ \Rightarrow y - \frac{2a}{m} = m \left( x - \frac{a}{m^2} \right)\]

\[ \Rightarrow \frac{my - 2a}{m} = m\left( \frac{m^2 x - a}{m^2} \right)\]

\[ \Rightarrow my - 2a = m^2 x - a\]

\[ \Rightarrow m^2 x - my + a = 0\]

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

\[y - y_1 = \frac{1}{\text { Slope of tangent}} \left( x - x_1 \right)\]

\[ \Rightarrow y - \frac{2a}{m} = \frac{- 1}{m}\left( x - \frac{a}{m^2} \right)\]

\[ \Rightarrow \frac{my - 2a}{m} = \frac{- 1}{m}\left( \frac{m^2 x - a}{m^2} \right)\]

\[ \Rightarrow m^3 y - 2a m^2 = - m^2 x + a\]

\[ \Rightarrow m^2 x + m^3 y - 2a m^2 - a = 0\]