Question

If the tangent to the curve x = a t², y = 2 at is perpendicular to x-axis, then its point of contact is

(a) (a, a)
(b) (0, a)
(c) (0, 0)
(d) (a, 0)

Answer 1

(c) (0, 0)

Let the required point be (x₁, y₁).

$$\begin{gathered} \mathrm{Since},\mathrm{the}\mathrm{point}\mathrm{lies}\mathrm{on}\mathrm{the}\mathrm{curve}. \\ \mathrm{Hence},x_1=a{t}^2\mathrm{and}{y}_1=2at \\ \mathrm{Now},x=a{t}^2\mathrm{and}y=2at \\ \Rightarrow \frac{dx}{dt}=2at\mathrm{and}\frac{dy}{dt}=2a \\ \Rightarrow \frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}=\frac{2a}{2at}=\frac{1}{t}=\frac{2a}{y} \\ \text{Slope of the tangent =}{\left(\frac{dy}{dx}\right)}_{\left(x_1,y_1\right)}\text{=}\frac{2a}{y_1} \\ \text{It is given that the tangent is perpendicular to the}\mathit{\text{y}}\text{-axis.} \\ \text{It means that it is parallel to the}\mathit{\text{x}}\text{-axis.} \\ \text{∴ Slope of the tangent = Slope of the}\mathit{\text{x}}\text{-axis} \\ \frac{2a}{y_1}=0 \\ \Rightarrow a=0 \\ \mathrm{Now}\text{,} \\ {x}_1=a{t}^2=0\mathrm{and}{y}_1=2at=0 \\ \therefore \left(x_1,y_1\right)=\left(0,0\right) \end{gathered}$$