CUET Preparation Today
| The equation of the tangent at the point \(P(t)\), where \(t\) is any parameter to the parabola \(y^{2}=4ax\) is |
\(y=xt+at^{2}\) \(yt=x+at^{2}\) \(y=tx\) \(y=tx+\frac{a}{t}\) |
| \(yt=x+at^{2}\) |
| 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, \(yt=x+at^{2}\) |
