Question

If the normals drawn to the hyperbola $xy=4$ 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. $\frac{-16b}{a}$
• B. $\frac{-16a}{b}$
• C. $\frac{4b}{a}$
• D. $\frac{4a}{b}$

Answer 1

Answer: (B)

The equation of normal to the given hyperbola $xy=4$ at point $\left(2t,\frac{2}{t}\right)$ is
$2t^4=x{t}^3+yt-2=0\dots (i)$
$\because$ Normal (i) passes through point $(a,b)$, so
$2t^4-a{t}^3+bt-2=0$, the equation having roots are $\frac{{\alpha }_1}{2},\frac{{\alpha }_2}{2},\frac{{\alpha }_3}{2}$ and $\frac{{\alpha }_4}{2}$, so
$\frac{1}{2}\left({\alpha }_1+{\alpha }_2+{\alpha }_3+{\alpha }_4\right)=\frac{a}{2}$
$\Rightarrow {\alpha }_1+{\alpha }_2+{\alpha }_3+{\alpha }_4=a$
$\sum \left(\frac{{\alpha }_1{\alpha }_2}{4}\right)=0$
$\sum \frac{{\alpha }_1{\alpha }_2{\alpha }_3}{8}=\frac{-b}{2}\text{ and }\frac{{\alpha }_1{\alpha }_2{\alpha }_3{\alpha }_4}{16}=-1$
$\because {\beta }_1+{\beta }_2+{\beta }_3+{\beta }_4=\frac{4}{{\alpha }_1}+\frac{4}{{\alpha }_2}+\frac{4}{{\alpha }_3}+\frac{4}{{\alpha }_4}$
$=4\frac{\Sigma {\alpha }_1{\alpha }_2{\alpha }_3}{{\alpha }_1{\alpha }_2{\alpha }_3{\alpha }_4}$
$=4\frac{-b/2\times 8}{-16}=b$
$\therefore \frac{\left({\alpha }_1+{\alpha }_2+{\alpha }_3+{\alpha }_4\right)}{{\beta }_1+{\beta }_2+{\beta }_3+{\beta }_4}\left({\alpha }_1{\alpha }_2{\alpha }_3{\alpha }_4\right)$
$=\frac{a(-16)}{b}=-16\frac{a}{b}$