The equation of the tangent at the point \(P(t)\), where \(t\) is any parameter to the parabola \(y^{2}=4ax\) is
Answer & explanation
Correct answer: option 2
Given, \(y^{2}=4ax\) So, \(\frac{dy}{dx}=\frac{2a}{y}\hspace{7cm}\)
\(\begin{aligned}n&=\left(\frac{dy}{dx}\right)_{(at^{2},2at)}\\ &=\frac{1}{t}\end{aligned}\hspace{7cm}\)
Equation: \(y-2at=\frac{1}{t}(x-at^{2})\) So, \(yt=x+at^{2}\)