Question

The curve orthogonal to the curve x² + 2xy – y² = k would be:
1. x² + 2xy – y² = k
2. x² – 2xy + y² = k
3. x² – 2xy – y² = k
4. None of these

Answer 1

Correct Answer - Option 3 : x² – 2xy – y² = k

The given curve is:

x² + 2xy – y² = k

\(2x + 2y + 2x\frac{{dy}}{{dx}} - 2y\frac{{dy}}{{dx}} = 0\) 

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

\(\frac{{dy}}{{dx}} = \frac{{y + x}}{{y - x}}\) 

Replacing \(\frac{{dy}}{{dx}}\) by \(\frac{{ - dx}}{{dy}}\) , we get:

\(\frac{{ - dx}}{{dy}} = \frac{{y + x}}{{y - x}}\) 

\(\frac{{dx}}{{dy}} = \frac{{x + y}}{{x - y}}\) 

\(\frac{{dy}}{{dx}} = \frac{{x - y}}{{x + y}}\)       ---(1)

This is a homogenous equation

i.e. y = υx

\(\frac{{dy}}{{dx}} = \upsilon + x\frac{{d\upsilon }}{{dx}}\)         ---(2)

From (1) and (2), we get:

\(\upsilon + x\frac{{d\upsilon }}{{dx}} = \frac{{\upsilon - \upsilon x}}{{\upsilon + \upsilon x}}\) 

\(x\frac{{d\upsilon }}{{dx}} = \frac{{1 - \upsilon }}{{1 + \upsilon }} - \upsilon \) 

= \(\frac{{1 - 2\upsilon - {\upsilon ^2}}}{{1 + \upsilon }}\)

Integrating both sides, we get:

\(\smallint \frac{{1 + \upsilon }}{{1 - 2\upsilon - {\upsilon ^2}}}d\upsilon = \smallint \frac{{dx}}{x}\) 

\(\frac{{ - 1}}{2}\smallint \frac{{ - 2\left( {1 + \upsilon } \right)}}{{1 - 2\upsilon - {\upsilon ^2}}}d\upsilon = \smallint \frac{{dx}}{x}\) 

 log (1 – 2υ – υ²) = -2 log x + log c

\({x^2} - 2xy - {y^2} = c\) 

The correct answer is:

x² – 2xy – y² = k