Question

Statement I The required condition, for the curves $2 x=y^2$ and $2 x y=K$ to intersect orthogonally, is $K^2=6$.
Statement II $\left(\frac{d y}{d x}\right)_{\text {1st curve }} \times\left(\frac{d y}{d x}\right)_{2 \text { nd curve }}=-1$
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, $2 x=y^2 .....$(i)
and $2 x y=k .....$(ii)
On differentiating Eq. (i), we get
$2=2 y \frac{d y}{d x} \Rightarrow m_1=\frac{d y}{d x}=\frac{1}{y}$
On differentiating Eq. (ii), we get
$ y+x\left(\frac{d y}{d x}\right) =0$
$\therefore m_2 =\frac{d y}{d x}=\frac{-y}{x}$
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.