Question

Without changing the direction of axes where the origin is to be shifted so that the equation $\left(a\right)x^2+y^2+4x-6y-3=0$ transforms to the form $x^2+y^2=a^2\left(b\right)y^2-3x+4y+13=0$ transforms to the form $y^2=ax?$ Find the transformed equation in each case.

Answer 1

Let the origin be shifted to $(h,k)$

$\left(a\right)$ $\Rightarrow x=X+h,y=Y+k$

Given equation $\to x^2+y^2+4x-6y-3=0$

$\Rightarrow (X+h{)}^2+(Y+k{)}^2+4(X+h)-6(Y+k)-3=0$

$\Rightarrow X^2+h^2+2hX+Y^2+k^2+2kY+4X+4h-6Y-6k-3=0$

$\Rightarrow X^2+Y^2+(2h+4)X+(2k-6)Y+(h^2+k^2+4h-6k-3)=0$

Given that equation is transformed as $X^2+Y^2=a^2$

Therefore, $\Rightarrow 2h+4=0\Rightarrow h=-2$

$\Rightarrow 2k-6=0\Rightarrow k=3$

$\Rightarrow h^2+k^2+4h-6k-3=-a^2$

$\Rightarrow (-2{)}^2+3^2+4(-2)-6(3)-3=-a^2$

$\Rightarrow a^2=16$

Thus, the transformed equation is $x^2+y^2=16$

$\left(b\right)$ $\Rightarrow x=X+h,y=Y+k$

Given equation $\to y^2-3x+4y+13=0$

$\Rightarrow (Y+k{)}^2-3(X+h)+4(Y+k)+13=0$

$\Rightarrow Y^2+k^2+2kY-3X-3h+4Y+4k+13=0$

$\Rightarrow Y^2-3X+(2k+4)Y+(k^2-3h+4k+13)=0$

Given that equation is transformed as $Y^2=aX$

Therefore, $\Rightarrow 2k+4=0\Rightarrow k=-2$

$\Rightarrow k^2-3h+4k+13=0$

$\Rightarrow (-2{)}^2-3h+4(-2)+13=0$

$\Rightarrow 3h=9\Rightarrow h=3$

Thus, the transformed equation is $y^2=-3x$