# 2x^2+Cy^2+Dx+Ey+F=0 represents a conic. what is the value of C for a hyperbola?

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### 1 Answer

You need to remember what the equation of hyperbola is such that:

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

You need to complete the squares in the equation provided by the problem to make it looks like equation of hyperbola such that:

`(x^2 + Dx/2 + (-D/4)^2)- (Cy^2/2 + Ey/2 + (-Esqrt2/(2sqrt C))^2) + F/2 - (-D/4)^2 - (-Esqrt2/(2sqrt C))^2= 0`

`(x - D/4)^2 - (Cy/sqrt2 - Esqrt2/(2sqrtC))^2 = (-D/4)^2+ (-Esqrt2/(2sqrt C))^2 - F/2`

You need to remember that `D^2/16 + E/(2C) - F/2=1 =gt D^2*C + 8E - 8FC = 16C `

You need to solve for C the equation `D^2*C + 8E - 8FC = 16C,` hence you need to isolate the terms containing C to the left side such that:

`C(D^2 - 8F - 16) = -8E =gt C = (-8E)/(D^2 - 8F - 16)`

**Hence, evaluating C for the conic to express a hyperbola yields `C = (-8E)/(D^2 - 8F - 16).` **