Question

the center of the equation below is (h,k), find h+k.

Answer 1

Answer:

The value of h+k is;

$h+k=-3$

Explanation:

Given the equation;

$81x^2-49y^2+324x-98y+4244=0$

Solving to derive the center of the function;

$\frac{(y-k)^2}{b^2}-\frac{(x-h)^2}{a^2}=1$

Simplifying;

$\begin{gathered} 81x^2-49y^2+324x-98y+4244=0 \\ 81x^2+324x-49y^2-98y+4244=0 \\ 81x^2+324x+324-49y^2-98y-49=-4244+324-49 \\ (81x^2+324x+324)-(49y^2+98y+49)_{}=-4244+324-49 \\ 81(x^2+4x+4)-49(y^2+2y+1)=-3969 \end{gathered}$

Dividing through by -3969;

$\begin{gathered} \frac{81(x^2+4x+4)}{-3969}-\frac{49(y^2+2y+1)}{-3969}=\frac{-3969}{-3969} \\ \frac{(y+1)^2}{81}-\frac{(x+2)^2}{49}=1 \\ \frac{(y-(-1))^2}{9^2}-\frac{(x-(-2))^2}{7^2}=1 \end{gathered}$

So;

$\begin{gathered} h=-2 \\ k=-1 \\ (h,k)=(-2,-1) \end{gathered}$

Therefore, the value of h+k is;

$\begin{gathered} h+k=-2+(-1) \\ h+k=-2-1 \\ h+k=-3 \end{gathered}$