Question

By shifting the origin to a suitable point O' (h, k) axes remaining parallel, reduce the equation $4x^2+4y^2+16x-18y+24=0$ to the form $\frac{x^2}{a^2}+\frac{y^2}{b^2}=2(a>0,b>0)$, find O'(h, k) and the values of a and b.

Answer 1

Given : $4x^2+4y^2+16x-18y+24=0$

divide by 4

$x^2+y^2+4x-\frac{9}{2}y+6=0$

$x^2+4x+4-4+y^2-\frac{9}{2}y+\frac{81}{16}-\frac{81}{16}+6=0$

${(x+2)}^2+{(y-\frac{9}{4})}^2+2-\frac{81}{16}=0$

${(x+2)}^2+{(y-\frac{9}{4})}^2=\frac{49}{16}$

$\frac{{(x+2)}^2}{\left(\frac{49}{32}\right)}+\frac{(y-9/4{)}^2}{\left(\frac{49}{32}\right)}=2$

$\frac{{(x+2)}^2}{{\left(\frac{7}{4\sqrt{2}}\right)}^2}+\frac{{(y-9/4)}^2}{{\left(\frac{7}{4\sqrt{2}}\right)}^2}=2$

$\frac{{[x-(-2\left)\right]}^2}{{\left(\frac{7}{4\sqrt{2}}\right)}^2}+\frac{{[y-(+9/4\left)\right]}^2}{{\left(\frac{7}{4\sqrt{2}}\right)}^2}=2$

$\therefore 0^1(h,k)=0^1(-2,9/4)$

$a=b=\frac{7}{4\sqrt{2}}$