If the normals drawn to the hyperbola \(x y=4\) at \(\left(\alpha_i, \beta_i\right)(i=1,2,3,4)\) are…
If the normals drawn to the hyperbola \(x y=4\) at \(\left(\alpha_i, \beta_i\right)(i=1,2,3,4)\) are concurrent at the point \((a, b)\), then
\(\frac{\left(\alpha_1+\alpha_2+\alpha_3+\alpha_4\right)}{\left(\beta_1+\beta_2+\beta_3+\beta_4\right)}\left(\alpha_1 \alpha_2 \alpha_3 \alpha_4\right)=\)
\(\frac{-16 b}{a}\)
\(\frac{-16 a}{b}\)
\(\frac{4 b}{a}\)
\(\frac{4 a}{b}\)
Solution
The equation of normal to the given hyperbola \(x y=4\) at point \(\left(2 t, \frac{2}{t}\right)\) is
\(2 t^4-x t^3+y t-2=0\) ...(i)
\(\because\) Normal (i) passes through point \((a, b)\), so \(2 t^4-a t^3+b t-2=0\), the equation having roots are \(\frac{\alpha_1}{2}, \frac{\alpha_2}{2}, \frac{\alpha_3}{2}\) and \(\frac{\alpha_4}{2}\), so
\(\begin{gathered}
\frac{1}{2}\left(\alpha_1+\alpha_2+\alpha_3+\alpha_4\right)=\frac{a}{2} \Rightarrow \alpha_1+\alpha_2+\alpha_3+\alpha_4=a \\
\sum\left(\frac{\alpha_1 \alpha_2}{4}\right)=0 \\
\sum \frac{\alpha_1 \alpha_2 \alpha_3}{8}=\frac{-b}{2} \text { and } \frac{\alpha_1 \alpha_2 \alpha_3 \alpha_4}{16}=-1 \\
\because \beta_1+\beta_2+\beta_3+\beta_4=\frac{4}{\alpha_1}+\frac{4}{\alpha_2}+\frac{4}{\alpha_3}+\frac{4}{\alpha_4}
\end{gathered}\)
\(=4 \frac{\Sigma \alpha_1 \alpha_2 \alpha_3}{\alpha_1 \alpha_2 \alpha_3 \alpha_4}\)
\(=4 \frac{-b / 2 \times 8}{-16}=b\)
\(\begin{aligned}
\frac{\left.\therefore \alpha_1+\alpha_2+\alpha_3+\alpha_4\right)}{\beta_1+\beta_2+\beta_3+\beta_4} & \left(\alpha_1 \alpha_2 \alpha_3 \alpha_4\right) \\
= & \frac{a(-16)}{b}=-16 \frac{a}{b}
\end{aligned}\)
Hence, option (b) is correct.