Question

x^2 +4x= - (y^2 -6)

write the conic equation in standard form

Answer 1

Step-by-step explanation:

${x}^{2} + 4x = - {y}^{2} + 6$

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

Standard Form of Conic is

$ax {}^{2} + bxy + c{y}^{2} + dx + ey + f = 0$

In this scenario, a is 1. c is 1 d is 4 and f is -6

Since a=c, we will get a circle.

Standard form of a circle is

$(x - h) {}^{2} + (y - k) {}^{2} = {r}^{2}$

where (h,k) is the center and r is the radius.

Step 1: Convert x^2+4x+y^2-6 into circle.

${x}^{2} + 4x + {y}^{2} - 6$

Add 6 to both sides

${x}^{2} + 4x + {y}^{2} = 6$

Complete the square with respect to x variable

Divide 4 by 2 and square it, then add that to both sides

$( \frac{4}{2} ) {}^{2} = 4$

So add four to both sides

${x}^{2} + 4x + 4 + {y}^{2} = 6 + 4$

Remebrr that a perfect square is

${a}^{2} + 2ab + {b}^{2} = (a + b) {}^{2}$

So we get

$(x + 2) {}^{2} + {y}^{2} = 10$

That is our standard equation.

Look at the graph above.