Respuesta :
Answer:
[tex](-\frac{85}{28},2)[/tex]
Step-by-step explanation:
We are asked to find the focus of the parabola that has a vertex [tex](-3,2)[/tex] opens horizontally and passes through the point [tex](-10,1)[/tex].
We know that equation of a horizontal parabola is [tex](y-k)^2=4p(x-h)[/tex], where p is not equal to zero.
[tex](h,k)[/tex] = Vertex of parabola.
Focus: [tex](h+p,k)[/tex]
Upon substituting the coordinates of vertex and given point in equation of parabola, we will get:
[tex](1-2)^2=4p(-10-(-3))[/tex]
[tex](-1)^2=4p(-10+3)[/tex]
[tex]1=4p(-7)[/tex]
[tex]1=-28p[/tex]
[tex]\frac{1}{-28}=\frac{-28p}{-28}[/tex]
[tex]-\frac{1}{28}=p[/tex]
Focus: [tex](h+p,k)[/tex]
[tex](-3+(-\frac{1}{28}),2)[/tex]
[tex](-3-\frac{1}{28},2)[/tex]
[tex](\frac{-3*28}{28}-\frac{1}{28},2)[/tex]
[tex](\frac{-84}{28}-\frac{1}{28},2)[/tex]
[tex](\frac{-84-1}{28},2)[/tex]
[tex](\frac{-85}{28},2)[/tex]
[tex](-\frac{85}{28},2)[/tex]
Therefore, the focus of the parabola is at point [tex](-\frac{85}{28},2)[/tex].
