If \(x-y=1\) is a tangent to the hyperbola \(\frac{x^{2}}{4}-\frac{y^{2}}{3}=1\), then the point of contact is

Given, \(\quad x-y=1\) and \(\quad \frac{x^{2}}{4}-\frac{y^{2}}{3}=1\)
Put \(y=x-1\) from Eq. (i) in Eq. (ii), we get
\[
\begin{aligned}
& & \frac{x^{2}}{4}-\frac{(x-1)^{2}}{3} &=1 \\
\Rightarrow & & \frac{3 x^{2}-4(x-1)^{2}}{12} &=1 \\
\Rightarrow & & 3 x^{2}-4\left(x^{2}+1-2 x\right) &=12 \\
\Rightarrow & & 3 x^{2}-4 x^{2}-4+8 x &=12 \\
\Rightarrow & &-x^{2}+8 x-16 &=0 \\
\Rightarrow & & x^{2}-8 x+16 &=0 \\
\Rightarrow & &x-4)^{2} &=0 \\
\Rightarrow & & x &=4,4
\end{aligned}
\]
Put \(x=4\) in Eq. (i), we get
\[
\begin{aligned}
\Rightarrow \quad 4-y &=1 \\
y &=3 \\
\therefore \quad \text { The point of contact } &=(4,3)
\end{aligned}
\]