Question

Show that the curves $4x=y^2$ and $4xy=k$ cut at right angles, if $k^2=512$.

Answer 1

The given curves are,

$\Rightarrow$ $4x=y^2$ ----- ( 1 )

$\Rightarrow$ $4xy=k$ ----- ( 2 )

We have to prove that two curves cut at right angles if $k^2=512$

Now,

$4x=y^2$

Differentiating both sides w.r.t.$x,$ we get

$\Rightarrow$ $4=2y.\frac{dy}{dx}$

$\Rightarrow$ $\frac{dy}{dx}=\frac{2}{y}$

$\Rightarrow$ $m_1=\frac{2}{y}$ ----- ( 3 )

$4xy=k$

Differentiating both sides w.r.t.$x,$ we get

$\Rightarrow$ $4(1\times y+x\frac{dy}{dx})=0$

$\Rightarrow$ $y+x\frac{dy}{dx}=0$

$\Rightarrow$ $\frac{dy}{dx}=\frac{-y}{x}$

$\Rightarrow$ $m_2=\frac{-y}{x}$ ----- ( 4 )

It is given that two curves intersect orthogonally.

$\therefore$ $m_1.m_2=-1$

$\Rightarrow$ $\frac{2}{y}\times \frac{-y}{x}=-1$ [ From ( 3 ) and ( 4 ) ]

$\Rightarrow$ $\frac{-2}{x}=-1$

$\Rightarrow$ $x=2$

Now,

$4xy=k$

$\Rightarrow$ $\left(y^2\right)y=k$ [ Since, $4x=y$ ]

$\Rightarrow$ $y^3=k$

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

Substituting $y=k^{\frac{1}{3}}$ in equation ( 1 ) we get,

$\Rightarrow$ $4x={\left(k^{\frac{1}{3}}\right)}^2$

$\Rightarrow$ $4\times 2=k^{\frac{2}{3}}$

$\Rightarrow$ $8=k^{\frac{2}{3}}$

$\Rightarrow$ $k^2=(8{)}^3$ [ Taking cube on both sides ]

$\Rightarrow$ $k^2=512$