3x + 4y – 7 = 0 is normal to$4{\mathrm{x}}^2 – 3{\mathrm{y}}^2 = 1$ at the point

$$\begin{gathered} Given hyperbola 4x^2-3y^2=1 \\ \Rightarrow \frac{{\mathrm{x}}^2}{{\displaystyle \frac{1}{4}}}-\frac{{\mathrm{y}}^2}{{\displaystyle \frac{1}{3}}}=1 \\ \Rightarrow {\mathrm{a}}^2=\frac{1}{4}, {\mathrm{b}}^2=\frac{1}{3} \\ \mathrm{equation} \mathrm{of} \mathrm{normal} \mathrm{at} \mathrm{P}({\mathrm{x}}_1, {\mathrm{y}}_1)\mathrm{is} \\ \frac{{\mathrm{a}}^2\mathrm{x}}{{\mathrm{x}}_1}+\frac{{\mathrm{b}}^2\mathrm{y}}{{\mathrm{y}}_1}={\mathrm{a}}^2+{\mathrm{b}}^2 \\ \Rightarrow \frac{\mathrm{x}}{4{\mathrm{x}}_1}+\frac{\mathrm{y}}{3{\mathrm{y}}_1}-\frac{7}{12}=0 — ① \\ \mathrm{Given} \mathrm{normal} 3\mathrm{x}+4\mathrm{y}-7=0 — ② \\ \mathrm{since} ① \& ② \mathrm{represents} \mathrm{same} \mathrm{lin}e \mathrm{then} \\ \Rightarrow \frac{3}{{\displaystyle \frac{1}{4{\mathrm{x}}_1}}}=\frac{4}{{\displaystyle \frac{1}{3{\mathrm{y}}_1}}}=\frac{\cancel{-} 7}{\cancel{-} {\displaystyle \frac{7}{12}}} \\ \Rightarrow 12{\mathrm{x}}_1=12{\mathrm{y}}_1 =12 \\ \Rightarrow ({\mathrm{x}}_1, {\mathrm{y}}_1)=(1, 1) \end{gathered}$$