Question

The equation of normal to the hyperbola $xy=4$ which is parallel to the line $2x-y=5$ is/are

• A. $2\sqrt{2}x+\sqrt{2}y=6$
• B. $-2\sqrt{2}x-\sqrt{2}y=6$
• C. $-2\sqrt{2}x+\sqrt{2}y=6$
• D. $2\sqrt{2}x-\sqrt{2}y=6$

Answer 1

Answer: (C), (D)

$2\sqrt{2}x-\sqrt{2}y=6$

Given, $xy=4\Rightarrow c=2$
Line: $2x-y=5\Rightarrow y=2x-5$
Slope of the line $=2$
Hence, Slope of the normal $=2$
Let the normal is drawn at $(x_1,y_1)$
Equation of normal to the hyperbola $xy=c^2$ at the point P$(x_1,y_1)$ is $x{x}_1-y{y}_1={x_1}^2-{y_1}^2.$
Slope of the normal: $\frac{x_1}{y_1}=2\Rightarrow x_1=2y_1.$
P$(x_1,y_1)$ satsfies $xy=4$
$\Rightarrow x_1{y}_1=4$
$\Rightarrow 2y_1{y}_1=4$
$\Rightarrow y_1=\pm \sqrt{2}$
$x_1=2\times \pm \sqrt{2}=\pm 2\sqrt{2}$
Hence, the point through which normal drawn are
$(2\sqrt{2},\sqrt{2})\text{and}(-2\sqrt{2},-\sqrt{2}).$
For point $(2\sqrt{2},\sqrt{2})$,
Equation of normal: $x2\sqrt{2}-y\sqrt{2}={\left(2\sqrt{2}\right)}^2-{\left(\sqrt{2}\right)}^2$
$\Rightarrow 2\sqrt{2}x-\sqrt{2}y=6$

For point $(-2\sqrt{2},-\sqrt{2})$,
Equation of normal:
$x(-2\sqrt{2})-y(-\sqrt{2})={(-2\sqrt{2})}^2-{(-\sqrt{2})}^2$
$\Rightarrow -x2\sqrt{2}+y\sqrt{2}=8-2$
$\Rightarrow -2\sqrt{2}x+\sqrt{2}y=6.$