Question

Find the standard form of the equation for the conic section represented by x^2 + 10x + 6y = 47.

Answer 1

The standard form of the equation for the conic section represented by
`x^2\:+\:10x\:+\:6y\:=\:47` is:

`4\left(-(3)/(2)\right)\left(y-12\right)=\left(x-\left(-5\right)\right)^2`

Explanation:

We know that:

`4p\left(y-k\right)=\left(x-h\right)^2` is the standard equation for an up-down facing Parabola with vertex at (h, k), and focal length |p|.

Given the equation

`x^2\:+\:10x\:+\:6y\:=\:47`

Rewriting the equation in the standard form

`4\left(-(3)/(2)\right)\left(y-12\right)=\left(x-\left(-5\right)\right)^2`

Thus,

The vertex (h, k) = (-5, 12)

Please also check the attached graph.

Therefore, the standard form of the equation for the conic section represented by
`x^2\:+\:10x\:+\:6y\:=\:47` is:

`4\left(-(3)/(2)\right)\left(y-12\right)=\left(x-\left(-5\right)\right)^2`

where

vertex (h, k) = (-5, 12)