The equation of the tangent to the parabola $y^2=8 x$, which is parallel to the line $4 x-y+3=0$ is
- $2 x-8 y+1=0$
- $8 x-2 y+1=0$
- $8 x+2 y+1=0$
- $2 x-8 y-1=0$
Solution
Differentiating w.r.t. $x$, we get $\begin{aligned} & 2 y \frac{\mathrm{~d} y}{\mathrm{~d} x}=8 \\ & \Rightarrow \frac{\mathrm{~d} y}{\mathrm{~d} x}=\frac{4}{y} \end{aligned}$
Slope of the line $4 x-y+3=0$ is 4 . Since the tangent is parallel to $4 x-y+3=0$, their slopes are equal $\begin{aligned} \therefore \quad & \frac{4}{y}=4 \\ & \Rightarrow y=1 \end{aligned}$
When $y=1, x=\frac{1}{8}$ $\therefore \quad$ Equation of the tangent at $\left(\frac{1}{8}, 1\right)$ is $\begin{aligned} & y-1=4\left(x-\frac{1}{8}\right) \\ & \Rightarrow 2 y-2=8 x-1 \\ & \Rightarrow 8 x-2 y+1=0 \end{aligned}$
Asked in: MHT CET 2024 (15 May Shift 1)