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 $xy=\frac{a^2}{2}={\left(\frac{a}{\sqrt{2}}\right)}^2=c^2$ (say). Since $x=ct,y=\frac{c}{t}$ satisfies $xy=c^2$
$\therefore (x,y)=\left(ct,\frac{c}{t}\right)(t\ne c)$ is called a 't' point on the rectangular hyperbola.
If $t_1$ and $t_2$ are the roots of the equation $x^2-4x+2=0$, then the point of intersection of tangents at '$t_1$' and '$t_2$ ' on $xy=c^2$ is -

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