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

$$\begin{gathered} \mathrm{Since},\left(x_1,y_1\right)\mathrm{lies}\mathrm{on}\mathrm{the}\mathrm{curve}.\mathrm{Therefore}, \\ {x_1}^2+3y_1-3=0...\left(1\right) \\ \mathrm{Now},x^2+3y-3=0 \\ \Rightarrow 2x+3\frac{dy}{dx}=0 \\ \Rightarrow \frac{dy}{dx}=\frac{-2x}{3} \\ \text{Slope of tangent,}\mathit{\text{m}}\text{=}{\left(\frac{dy}{dx}\right)}_{\left(x_1,y_1\right)}\text{=}\frac{-2x_1}{3} \\ \text{Given that tangent is parallel to the line, So} \\ \text{Slope of}\mathrm{tangent}\text{,}\mathit{\text{m}}\text{= slope of the given line} \\ \frac{-2x_1}{3}=4 \\ \Rightarrow x_1=-6 \\ 36+3y_1-3=0\text{(From (1))} \\ \Rightarrow 3y_1=-33 \\ \Rightarrow y_1=-11 \\ \left(x_1,y_1\right)=\left(-6,-11\right) \\ \text{Equation of}\mathrm{tangent}\text{is,} \\ y-y_1=m\left(x-x_1\right) \\ \Rightarrow y+11=4\left(x+6\right) \\ \Rightarrow y+11=4x+24 \\ \Rightarrow 4x-y+13=0 \end{gathered}$$