Question

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 $\frac{\left(\alpha_{1}+\alpha_{2}+\alpha_{3}+\alpha_{4}\right)}{\left(\beta_{1}+\beta_{2}+\beta_{3}+\beta_{4}\right)}\left(\alpha_{1} \alpha_{2} \alpha_{3} \alpha_{4}\right)=$

• A. $\frac{-16 b}{a}$
• B. $\frac{-16 a}{b}$
• C. $\frac{4 b}{a}$
• D. $\frac{4 a}{b}$

Answer 1

Answer: (B)

The equation of normal to the given hyperbola $x y=4$ at point $\left(2 t, \frac{2}{t}\right)$ is
$2 t^{4}=x t^{3}+y t-2=0\,\,\,\,\,\dots(i)$
$\because$ 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
$\frac{1}{2}\left(\alpha_{1}+\alpha_{2}+\alpha_{3}+\alpha_{4}\right)=\frac{a}{2}$
$ \Rightarrow \alpha_{1}+\alpha_{2}+\alpha_{3}+\alpha_{4}=a$
$\sum\left(\frac{\alpha_{1} \alpha_{2}}{4}\right)=0$
$\sum \frac{\alpha_{1} \alpha_{2} \alpha_{3}}{8}=\frac{-b}{2} \text { and } \frac{\alpha_{1} \alpha_{2} \alpha_{3} \alpha_{4}}{16}=-1$
$\because \,\beta_{1}+\beta_{2}+\beta_{3}+\beta_{4}=\frac{4}{\alpha_{1}}+\frac{4}{\alpha_{2}}+\frac{4}{\alpha_{3}}+\frac{4}{\alpha_{4}}$
$=4 \frac{\Sigma \alpha_{1} \alpha_{2} \alpha_{3}}{\alpha_{1} \alpha_{2} \alpha_{3} \alpha_{4}}$
$=4 \frac{-b / 2 \times 8}{-16}=b$
$\therefore \, \frac{\left(\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) $
$=\frac{a(-16)}{b}=-16 \frac{a}{b}$