Question

If we rotate the axes of the rectangular hyperbola $x^2-y^2=a^2$ through an angle $\pi / 4$ in the clockwise direction then the equation $x^2-y^2=a^2$ reduces to $x y=\frac{a^2}{2}=\left(\frac{a}{\sqrt{2}}\right)^2=c^2$ (say). Since $x=c t, y=\frac{c}{t}$ satisfies $x y=c^2$
$\therefore (x, y)=\left(c t, \frac{c}{t}\right)(t \neq c)$ is called a 't' point on the rectangular hyperbola.
If $t_1$ and $t_2$ are the roots of the equation $x^2-4 x+2=0$, then the point of intersection of tangents at '$ t_1$' and '$ t_2$ ' on $x y=c^2$ is -

• A. $\left(\frac{ c }{2}, 2 c \right)$
• B. $\left(2 c , \frac{ c }{2}\right)$
• C. $\left(\frac{c}{2}, c\right)$
• D. $\left( c , \frac{ c }{2}\right)$

Answer 1

Answer: (D)

Tangent of $xy = c ^2$ at $t _1 \& t _2$ are
$x+t_1^2 y =2 c t_1 \ldots(i)$
and $x+t_2^2 y =2 c t_2 \ldots \text { (ii) }$
on solving (i) & (ii) we get
$y=\frac{2 c}{t_1+t_2}=\frac{2 c}{4}, x=\frac{2 c_1 t_2}{t_1+t_2}=\frac{4 c}{4}$
$\therefore$ point of intersection is $\left(c, \frac{c}{2}\right)$.