Question

A rectangular hyperbola is drawn by rotating the axis of hyperbola $\frac{x^2}{16}-\frac{y^2}{16}=1$ in clockwise direction by ${45}^{\circ }.$ The equation of the latus rectum of the rectangular hyperbola is:

• A. $x+y=\pm 8$
• B. $x+y=\pm 2$
• C. $2x+y=\pm 8$
• D. $x+y=\pm 4$

Answer 1

The correct option is A $x+y=\pm 8$

Given, Hyperbola ;$\frac{x^2}{16}-\frac{y^2}{16}=1$
$\Rightarrow x^2-y^2=16$
Let the rectangular hyperbola be $xy=c^2$
As this hyperbola formed by rotation of
$\frac{x^2}{16}-\frac{y^2}{16}=1$
$\Rightarrow c^2=\frac{a^2}{2}=\frac{16}{2}=8$
Rectangular hyperbola be
$xy=8\Rightarrow c=2\sqrt{2}$
The equation of latus rectum to a rectangular hyperbola
$xy=c^2\text{is}\Rightarrow x+y=\pm 2\sqrt{2}c$
$\Rightarrow x+y=\pm \left(2\sqrt{2}\right)\times \left(2\sqrt{2}\right)$
$\Rightarrow x+y=\pm 8$
Hence, the correct answer is option a.