CUET Preparation Today
The area bounded by y2 = – 4x, and its latus-rectum is |
Not defined 1 sq. units 2/3 sq. units none of these |
none of these |
The required area = $\begin{vmatrix}-2\int\limits_{-1}^0(-4x)^{1/2}dx\end{vmatrix}$ $=\begin{vmatrix}-2×\frac{2}{3}\frac{(-4x)^{3/2}}{-4}|_{-1}^0\end{vmatrix}$ $=\begin{vmatrix}\frac{1}{3}[0-(4)^{3/2}]=\frac{8}{3}\end{vmatrix}$ Hence (D) is the correct answer. |
