Question

The straight lines represented by $(y-mx{)}^2=a^2(1+m^2)$ and $(y-nx{)}^2=a^2(1+n^2)$ forms
(where $mn\ne -1$ and $m\ne n$)

• A. square
• B. rhombus
• C. rectangle
• D. trapezium

Answer 1

The correct option is B rhombus

We have,
$(y-mx{)}^2=a^2(1+m^2)$ and $(y-nx{)}^2=a^2(1+n^2)$
$\Rightarrow y-mx=\pm a\sqrt{1+m^2}\text{and}y-nx=\pm a\sqrt{1+n^2}$
$\Rightarrow y=mx\pm a\sqrt{1+m^2}\text{and}y=nx\pm a\sqrt{1+n^2}$
Thus, the equations of the lines represented by the given equations are
$y=mx+a\sqrt{1+m^2}...\left(i\right)$
$y=mx-a\sqrt{1+m^2}...\left(ii\right)$
$y=nx+a\sqrt{1+n^2}...\left(iii\right)$
$y=nx-a\sqrt{1+n^2}...\left(iv\right)$

Clearly, lines $\left(i\right)$ and $\left(ii\right)$ are parallel and the distance between them is given by
$d_1=\left|\frac{a\sqrt{1+m^2}+a\sqrt{1+m^2}}{\sqrt{1+m^2}}\right|=|2a|$
Similarly, lines $\left(iii\right)$ and $\left(iv\right)$ are parallel and the distance between them is given by
$d_2=\left|\frac{a\sqrt{1+n^2}+a\sqrt{1+n^2}}{\sqrt{1+n^2}}\right|=|2a|$
Clearly, $d_1=d_2$

Hence the lines $\left(i\right),\left(ii\right),\left(iii\right)$ and $\left(iv\right)$ form a rhombus.