If the expression $$\frac{{{x^2}}}{{{y^2}}} + tx + \frac{{{y^2}}}{4}$$   is a perfect square, then the value of t is?

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