Question

To change $(3x+4y+5)-(2x+3y+4)\frac{dy}{dx}=0$ into homogeneous equation, origin is shifted to $(h,k)$ then$h+k=$

• A. $3$
• B. $1$
• C. $-2$
• D. $-1$

Answer 1

Answer: (A)

$3$

$\left(3x+4y+5\right)-\left(2x+2y+4\right)\frac{dy}{dx}=0$

$\Rightarrow \frac{dy}{dx}=\frac{3x+4y+5}{2x+3y+4}$

shifting origin to $(h,k)$ then we get $x=x+h,y=y+k$ where $x,y$ are in new coordinates system

$\begin{array}{c}\frac{dy}{dx}=\frac{3\left(x+h\right)+4\left(y+k\right)+5}{2\left(x+h\right)+3\left(y+k\right)+4}\\ \Rightarrow \frac{dy}{dx}=\frac{3x+4y+3h+4k+5}{2x+3y+2h+3k+4}\\ Solvingfor\left(h,k\right)sothat\\ 3h+4k+5=0-eq1\\ and2h+3k+4=0-eq2\\ Multiplyingeq1by2and2by3\\ 6h+8k+10=0-eq3\\ 6h+9k+12=0-eq4\\ subtractingeq4from3weget\\ -k-2=0\\ \Rightarrow k=-2\\ puttingk=-2ineq1\\ 3h+4k+5=0\\ \Rightarrow 3h-8+5=0\\ \Rightarrow 3h-3=0\\ h=1\\ therefore\\ h+k=1-2=-1\end{array}$