Question

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

Options

• (a, a)

• (0, a)

• (0, 0)

• (a, 0)

Answer 1

(0, 0)

 

Let the required point be (x₁, y₁).

\[\text { Since, the point lies on the curve } . \]

\[\text { Hence, } x_1 = a t^2 \text { and } y_1 = 2\text { at }\]

\[\text { Now }, x = a t^2 \text { and } y = 2\text { at }\]

\[ \Rightarrow \frac{dx}{dt} = 2\text { at  and } \frac{dy}{dt} = 2a\]

\[ \Rightarrow \frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{2a}{2at} = \frac{1}{t} = \frac{2a}{y}\]

\[\text { Slope of the tangent } = \left( \frac{dy}{dx} \right)_\left( x_1 , y_1 \right) =\frac{2a}{y_1}\]

\[\text { It is given that the tangent is perpendicular to the y-axis. }\]

\[\text { It means that it is parallel to thex-axis }.\]

\[\therefore \text { Slope of the tangent = Slope of the x-axis }\]

\[\frac{2a}{y_1} = 0\]

\[ \Rightarrow a = 0\]

\[\text { Now },\]

\[ x_1 = a t^2 = 0 \text { and } y_1 = 2\text { at }= 0\]

\[ \therefore \left( x_1 , y_1 \right) = \left( 0, 0 \right)\]