If the expression x^2y^2 + tx + y^24 is a perfect square, then the value of t is?

Answer: Option A

$$\frac{x^2}{y^2}+tx+\frac{y^2}{4}\left(\text{ Given}\right)$$
To make it a perfect square it should be in the form
$$\begin{array}{rl}& {\text{A}}^2\pm 2\text{AB}+{\text{B}}^2={\left(\text{A}\pm \text{B}\right)}^2\\ & ={\left(\frac{x}{y}\right)}^2\pm tx+{\left(\frac{y}{2}\right)}^2\\ & ={\text{A}}^2\pm 2\text{AB}+{\text{B}}^2\\ & \text{A}=\frac{x}{y}\text{, B}=\frac{y}{2}\&\text{2AB}=tx\\ & \text{So, }tx=\pm 2\times \frac{x}{y}\times \frac{y}{2}\\ & \Rightarrow tx=\pm x\\ & \Rightarrow t=\pm 1\end{array}$$