Question

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

Answer 1

The value of h+k is;

`h+k=-3`

Step-by-step explanation:

Given the equation;

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

Solving to derive the center of the function;

`((y-k)^2)/(b^2)-((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} (81(x^2+4x+4))/(-3969)-(49(y^2+2y+1))/(-3969)=(-3969)/(-3969) \\ ((y+1)^2)/(81)-((x+2)^2)/(49)=1 \\ ((y-(-1))^2)/(9^2)-((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}`