Question

Find the equation of the tangent to the curve x² + 3y − 3 = 0, which is parallel to the line y= 4x − 5 ?

Answer 1

Suppose (x₁, y₁) be the point of contact of tangent.
We can find the slope of the given line by differentiating the equation w.r.t  x
So, Slope of the line  = 4

\[\text { Since }, \left( x_1 , y_1 \right)\text {  lies on the curve . Therefore,} \]

\[ {x_1}^2 + 3 y_1 - 3 = 0 . . . \left( 1 \right)\]

\[\text { Now,} x^2 + 3y - 3 = 0\]

\[ \Rightarrow 2x + 3\frac{dy}{dx} = 0\]

\[ \Rightarrow \frac{dy}{dx} = \frac{- 2x}{3}\]

\[\text { Slope of tangent },m= \left( \frac{dy}{dx} \right)_\left( x_1 , y_1 \right) =\frac{- 2 x_1}{3}\]

\[\text { Given that tangent is parallel to the line, So }\]

\[\text { Slope of tangent, m = slope of the given line }\]

\[\frac{- 2 x_1}{3} = 4\]

\[ \Rightarrow x_1 = - 6\]

\[36 + 3 y_1 - 3 = 0...................[\text { From }(1)]\]

\[ \Rightarrow 3 y_1 = - 33\]

\[ \Rightarrow y_1 = - 11\]

\[\left( x_1 , y_1 \right) = \left( - 6, - 11 \right)\]

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

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

\[ \Rightarrow y + 11 = 4 \left( x + 6 \right)\]

\[ \Rightarrow y + 11 = 4x + 24\]

\[ \Rightarrow 4x - y + 13 = 0\]