Show that the curves

Solution:

Given:

Curves $2 x=y^{2} \ldots(1)$

$\& 2 x y=k \ldots(2)$

We have to prove that two curves cut at right angles if $k^{2}=8$

Now, Differentiating curves (1) \& (2) w.r.t x, we get

$\Rightarrow 2 x=y^{2}$

$\Rightarrow 2=2 y \cdot \frac{d y}{d x}$

$\Rightarrow \frac{d y}{d x}=\frac{1}{y}$

$\mathrm{m}_{1}=\frac{1}{\mathrm{y}} \ldots(3)$

$\Rightarrow 2 x y=k$

Differentiating above w.r.t $x$,

$\Rightarrow 2\left(1 \times y+x \frac{d y}{d x}\right)=0$

$\Rightarrow y+x \frac{d y}{d x}=0$

$\Rightarrow \frac{d y}{d x}=\frac{-y}{x}$

$\Rightarrow m_{2}=\frac{-y}{x} \ldots(4)$

Since $m_{1}$ and $m_{2}$ cuts orthogonally,

$\Rightarrow \frac{1}{y} \times \frac{-y}{x}=-1$

$\Rightarrow \frac{-1}{x}=-1$

$\Rightarrow x=1$

Now, Solving (1) \& (2), we get,

$2 x y=k \& 2 x=y^{2}$

$\Rightarrow\left(y^{2}\right) y=k$

$\Rightarrow y^{3}=k$

$\Rightarrow y=k^{\frac{1}{3}}$

Substituting $y=k^{\frac{1}{3}}$ in $2 x=y^{2}$, we get,

$\Rightarrow 2 x=\left(k^{\frac{1}{3}}\right)^{2}$

$\Rightarrow 2 \times 1=k^{\frac{2}{3}}$

$\Rightarrow k^{\frac{2}{3}}=2$

$\Rightarrow k^{2}=2^{3}$

$\Rightarrow k^{2}=8$