If the normals drawn to the hyperbola xy=4 at | vedclass

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(B) The equation of the normal to the hyperbola $xy=c^2$ (where $c^2=4$) at point $(ct, c/t)$ is $x t^3 - y t - c(t^4-1) = 0$. Substituting $c=2$,we g

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$\frac{-16a}{b}$

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(B) The equation of the normal to the hyperbola $xy=c^2$ (where $c^2=4$) at point $(ct, c/t)$ is $x t^3 - y t - c(t^4-1) = 0$. Substituting $c=2$,we get $x t^3 - y t - 2(t^4-1) = 0$,or $2t^4 - xt^3 + yt - 2 = 0$. Since the normal passes through $(a, b)$,we have $2t^4 - at^3 + bt - 2 = 0$. Let the roots be $t_1, t_2, t_3, t_4$. Then $\alpha_i = 2t_i$ and $\beta_i = 2/t_i$. From Vieta's formulas: $\sum t_i = a/2 \Rightarrow \sum \alpha_i = a$. $\sum t_i t_j t_k = -b/2$ and $t_1 t_2 t_3 t_4 = -1$. Note that $\alpha_1 \alpha_2 \alpha_3 \alpha_4 = 16(t_1 t_2 t_3 t_4) = 16(-1) = -16$. Also,$\sum \beta_i = \sum \frac{2}{t_i} = 2 \frac{\sum t_j t_k t_l}{t_1 t_2 t_3 t_4} = 2 \frac{-b/2}{-1} = b$. Thus,$\frac{\sum \alpha_i}{\sum \beta_i} (\alpha_1 \alpha_2 \alpha_3 \alpha_4) = \frac{a}{b} (-16) = -16 \frac{a}{b}$.

If the normals drawn to the hyperbola $xy=4$ at $(\alpha_i, \beta_i)$ for $i=1, 2, 3, 4$ are concurrent at the point $(a, b)$,then $\frac{(\alpha_1+\alpha_2+\alpha_3+\alpha_4)}{(\beta_1+\beta_2+\beta_3+\beta_4)}(\alpha_1 \alpha_2 \alpha_3 \alpha_4) =$

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