Answer:
The standard form of the equation for the conic section represented by [tex]x^2\:+\:10x\:+\:6y\:=\:47[/tex] is:
[tex]4\left(-\frac{3}{2}\right)\left(y-12\right)=\left(x-\left(-5\right)\right)^2[/tex]
Step-by-step explanation:
We know that:
[tex]4p\left(y-k\right)=\left(x-h\right)^2[/tex] is the standard equation for an up-down facing Parabola with vertex at (h, k), and focal length |p|.
Given the equation
[tex]x^2\:+\:10x\:+\:6y\:=\:47[/tex]
Rewriting the equation in the standard form
[tex]4\left(-\frac{3}{2}\right)\left(y-12\right)=\left(x-\left(-5\right)\right)^2[/tex]
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 [tex]x^2\:+\:10x\:+\:6y\:=\:47[/tex] is:
[tex]4\left(-\frac{3}{2}\right)\left(y-12\right)=\left(x-\left(-5\right)\right)^2[/tex]
where
vertex (h, k) = (-5, 12)
