Application of Derivatives — Class 12 Maths Solution

The given equation of curves are : $2x = {y^2}$
and $2xy = k$

$\Rightarrow$ $y = \frac{k}{{2x}}$ [from Eq.(ii)]

From Eq. (i), $2x = {\left( {\frac{k}{{2x}}} \right)^2}$
$\Rightarrow$ $8{x^3} = {k^2}$

$\Rightarrow$ ${x^3} = \frac{1}{8}{k^2}$

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

Therefore,$y = \frac{k}{{2x}} = \frac{k}{{2 \cdot \frac{1}{2}{k^{2/3}}}} = {k^{1/3}}$

Thus, we get point of intersection of curves which is $\left( {\frac{1}{2}{k^{2/3}},{k^{1/3}}} \right)$.

From Eqs. (i) and (ii),

$2 = 2y\frac{{dy}}{{dx}}$
and $2\left[ {x \cdot \frac{{dy}}{{dx}} + y \cdot 1} \right] = 0$
$\Rightarrow$ $\frac{{dy}}{{dx}} = \frac{1}{y}$

and $\left( {\frac{{dy}}{{dx}}} \right) = \frac{{ - 2y}}{{2x}} = - \frac{y}{x}$

$\Rightarrow$ ${\left( {\frac{{dy}}{{dx}}} \right)_{\left( {\frac{1}{2}{k^{2/3}},{k^{1/3}}} \right)}} = \frac{1}{{{k^{1/3}}}}$. [say ${m_1}$]

and ${\left( {\frac{{dy}}{{dx}}} \right)_{\left( {\frac{1}{2}{k^{2/3}},{k^{1/3}}} \right)}} = \frac{{ - {k^{1/3}}}}{{\frac{1}{2}{k^{2/3}}}} = - 2{k^{ - 1/3}}$ [say ${m_2}$]

Since, the curves intersect orthogonally.
i.e., ${m_1} \cdot {m_2} = - 1$

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

$\Rightarrow$ $- 2{k^{ - 2/3}} = - 1$

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

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

Therefore,${k^2} = 8$

which is the required condition.