• question_answer 14) The point of intersection of the tangents at the ends of the latus rectum of the parabola ${{\mathbf{y}}^{\mathbf{2}}}=\mathbf{4x}$ is A) $\left( 1,0 \right)$                      B) $\left( 0,1 \right)$          C) $\left( 0,-1 \right)$                     D) $\left( -1,0 \right)$

[d] The coordinate of extremities of the latus rectum of ${{y}^{2}}=4x$ be (1, 2) and$\left( 1,-2 \right)$. So, the equation of the tangents at (1, 2) and $\left( 1,-2 \right)$ be $2y=\frac{4\left( x+ \right)}{2}\Rightarrow y=x+1$                  ??...(1) $-2y=\frac{4\left( x+ \right)}{2}\Rightarrow -y=x+1$             ?.......(2) Solving equation (1) and (2), we have $2\left( x+1 \right)=0$ $x=-1$and $y=0$ Thus, required point be (? 1, 0). Hence, option [d] is correct