Question

Statement I The required condition, for the curves $2x=y^2$ and $2xy=K$ to intersect orthogonally, is $K^2=6$.
Statement II ${\left(\frac{dy}{dx}\right)}_{\text{1st curve }}\times {\left(\frac{dy}{dx}\right)}_{2\text{ nd curve }}=-1$
If first and second curves cut each other orthogonally.

• A. Statement I is true; statement II is true; statement II is a correct explanation for statement I
• B. Statement I is true; statement II is true; statement II is not a correct explanation for statement I
• C. Statement I is true; statement II is false
• D. Statement I is false, statement II is true

Answer 1

Answer: (D)

Given, $2x=y^2.....$(i)
and $2xy=k.....$(ii)
On differentiating Eq. (i), we get
$2=2y\frac{dy}{dx}\Rightarrow m_1=\frac{dy}{dx}=\frac{1}{y}$
On differentiating Eq. (ii), we get
$y+x\left(\frac{dy}{dx}\right)=0$
$\therefore m_2=\frac{dy}{dx}=\frac{-y}{x}$
For intersection point, solve Eqs. (i) and (ii), we get
$\frac{K}{y}=y^2\Rightarrow y^3=K\Rightarrow y=K^{1/3}$
$\therefore x=\frac{y^2}{2}=\frac{K^{2/3}}{2}.....$(iii)
If Eqs. (i) and (ii) cuts orthogonally,
Then, $m_1\times m_2=-1\Rightarrow \frac{1}{y}\times \frac{-y}{x}=-1\Rightarrow \frac{1}{x}=1\Rightarrow x=1$
$\Rightarrow \frac{K^{2/3}}{2}=1$
$\Rightarrow K^{2/3}=2$
[from Eq. (iii)]
Cubing on both sides, we get $K^2=8$ which is the required condition.