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If the tangent and normal at a point P on a rectangular hyperbola \mathrm{x y=c^2} cuts off intercepts \mathrm{ \lambda_1, \lambda_2} on the x-axis and \mathrm{ \mu_1, \mu_2} on the y-axis, then \mathrm{ \lambda_1 \lambda_2+\mu_1 \mu_2} is equal to

Option: 1

0


Option: 2

1


Option: 3

2


Option: 4

4


Answers (1)

best_answer

Let P be \mathrm{\left(c t, \frac{c}{t}\right) \, \, on \, \, x y=c^2.}

The equation of tangent at P is \mathrm{\frac{1}{2}\left(x \frac{c}{t}+y c t\right)=c^2}

The equation of normal at P is

\mathrm{ y-\frac{c}{t}=t^2(x-c t) }
The equations are rewritten as

\mathrm{ \frac{x}{2 c t}+\frac{y}{\frac{2 c}{t}}=1 }                              ..........(1)
And \mathrm{ x-\frac{y}{t^2}=c \frac{t^4-1}{t^3}}               ..........(2)
\mathrm{ \begin{aligned} \therefore \lambda_1 & =2 c t, \mu_1=\frac{2 c}{t} \\ \lambda_2 & =c \frac{t^4-1}{t^3}, \mu_2=-c \frac{t^4-1}{t} \end{aligned} }

\mathrm{\text { Hence, } \lambda_1 \lambda_2+\mu_1 \mu_2=\frac{2 c^2}{t^2}\left(t^4-1\right)-\frac{2 c^2}{t^2}\left(t^4-1\right)=0}
The answer is (a)

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Nehul

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