Question

Identify the type of conic represented by x2 + 4y2+2x-24y-63=0 and write the equation in standard form.....................................

Answer 1

To answer this question we will complete 2 perfect trinomials.

Notice that:

`x^2+4y^2+2x-24y-63=x^2+2x+4(y^2-6y)-63.`

Then:

`x^2+2x+4(y^2-6y)-63=0.`

Compleating the perfect trinomials we get:

`x^2+2x+4-4+4(y^2-6y+9-9)-63=0.`

Simplifying the above result we get:

`\begin{gathered} (x+2)^2-4+4(y-3)^2-36-63=0, \\ (x+2)^2+4(y-3)^2-103=0. \end{gathered}`

Adding 103 to the above result we get:

`(x+2)^2+4(y-3)^2=103.`

Dividing by 103 we get:

`((x+2)^2)/(103)+(4(y-3)^2)/(103)=1.`

We can rewrite the above equation as follows:

`((x+2)^2)/((√(103))^2)+((y-3)^2)/(((√(103))/(2))^2)=1.`

Now, notice that the above equation is as follows:

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

Therefore the given equation represents an ellipse.

Answer: The given equation represents an ellipse.

`((x+2)^2)/((√(103))^2)+((y-3)^2)/(((√(103))/(2))^2)=1.`