Question

Prove that the equation of tangent at the point (at^2 , 2at) on the parabola y^2 = 4ax is x - ty + at^2 = 0

Answer 1

Here the parabola given in the question is $y^2=4ax$ . So, to find the slope at the point $(at^2,2at)$ on the parabola we differentiate:-

$$y^2=4ax\\$$

Differentiating both sides:-

$$2y\frac{dy}{dx}=4a\\ \Rightarrow \frac{dy}{dx}=\frac{4a}{2y}=\frac{2a}{y}$$

So, slope at the required point:-

$$\frac{dy}{dx}|_{(at^2,2at)}=\frac{2a}{2at}=\frac{1}{t}$$

So, we know the point and the slope of the curve at that point. So the equation of the tangent is:-

$$(y-2at)=\frac{1}{t}(x-at^2)\\~\\ \Rightarrow yt-2at^2=x-at^2\\ \Rightarrow x-ty+at^2=0$$

Therefore, the equation of tangent at $(at^2,2at)$ on the parabola $y^2=4ax$ is:-

$$\fcolorbox{red}{cyan}{$\textcolor{black}{x-ty+at^2=0}$}$$