Question

If the normal to the rectangular hyperbola $xy=2$ at the point $P\left(2\right)$ meets the curve again at the point $Q$, then the coordinates of $P\text{and}Q$ are:

• A. $P(2\sqrt{2},\frac{1}{\sqrt{2}})$
  $Q(\frac{-1}{4\sqrt{2}},-8\sqrt{2})$
• B. $P(\sqrt{2},\frac{1}{\sqrt{2}})$
  $Q(\frac{-1}{4\sqrt{2}},-8\sqrt{2})$
• C. $P(2\sqrt{2},\frac{1}{\sqrt{2}})$
  $Q(\frac{1}{4\sqrt{2}},8\sqrt{2})$
• D. $P(2,\frac{1}{2})$
  $Q(\frac{-1}{4},-8)$

Answer 1

The correct option is A $P(2\sqrt{2},\frac{1}{\sqrt{2}})$

$Q(\frac{-1}{4\sqrt{2}},-8\sqrt{2})$
If the normal to the rectangular hyperbola $xy=c^2$ at the point $\left(t_1\right)$ meets the curve again at the point $\left(t_2\right)$, then $t_1^3{t}_2=-1$
Normal is drawn at Point $P\left(2\right)$
$\therefore t_1=2$
So, substitute $t_1=2\text{in}{t}_1^3{t}_2=-1$
$\Rightarrow 2^3{t}_2=-1$
$\Rightarrow t_2=\frac{-1}{8}$
$xy=c^2\Rightarrow c=\sqrt{2}$
Let the parametric coordinates of $P$ and $Q$ are $(c{t}_1,\frac{c}{t_1})\text{and}(c{t}_2,\frac{c}{t_2})$


$P(c{t}_1,\frac{c}{t_1})\equiv (\sqrt{2}\times 2,\frac{\sqrt{2}}{2})\equiv (2\sqrt{2},\frac{1}{\sqrt{2}})$

$Q(c{t}_2,\frac{c}{t_2})\equiv (\sqrt{2}\times \frac{-1}{8},\frac{\sqrt{2}}{\frac{-1}{8}})\equiv (\frac{-1}{4\sqrt{2}},-8\sqrt{2})$
Hence, the correct answer is option (a).