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

The center of the equation below is in the form (h,k), find h+k.-example-1
User Skunkfrukt
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The function given in the question is


9x^2+y^2+18x-12y-180=0

Rearrange the function above connecting similar terms


\begin{gathered} 9x^2+y^2+18x-12y-180=0 \\ 9x^2+18x+y^2-12y=180 \end{gathered}

Factorize the equation in brackets


\begin{gathered} 9x^2+18x+y^2-12y=180 \\ 9(x^2+2x)+1(y^2-12y)=180 \\ \end{gathered}

Solve the brackets using completing the square method but multiplying the coefficient of x and y by 1/2 and then square


\begin{gathered} 9(x^2+2x)+1(y^2-12y)=180 \\ 9(x^2+2x+(2*(1)/(2))^2)+1(y^2-12y+(-12*(1)/(2))^2=180+(2*(1)/(2))^2+(-12*(1)/(2))^2 \end{gathered}

By simplifying the equation above, we will have


\begin{gathered} 9(x^2+2x+(2*(1)/(2))^2)+1(y^2-12y+(-12*(1)/(2))^2=180+9(2*(1)/(2))^2+1(-12*(1)/(2))^2 \\ 9(x^2+2x+1)+1(y^2-12y+36)=180+9+36 \\ 9(x+1)^2+1(y-6)^2=225 \end{gathered}

Divide all through by 225


\begin{gathered} 9(x+1)^2+1(y-6)^2=225 \\ (9(x+1)^2)/(225)+(1(y-6)^2)/(225)=(225)/(225) \\ ((x+1)^2)/(25)+((y-6)^2)/(225)=1 \end{gathered}

The general formula for the equation of an ellipse is


((x-h)^2)/(a^2)+((y-k)^2)/(b^2)=1

Where the center of the ellipse is


(h,k)

By comparing coefficients, we will have that


\begin{gathered} x+1=x-h,y-6=y-k \\ x-x+h=-1,y-y+k=6 \\ h=-1,k=6 \end{gathered}

The centre of the ellipse is


\begin{gathered} (h,k)=(-1,6) \\ \text{hence,} \\ h+k=-1+6 \\ h+k=5 \end{gathered}

Therefore,

The value of h+k = 5

User Gonzalesraul
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