SOLUTION: Find the equation of the line satisfying the given conditions: Passes through the center of the circle (x+2)^2+(y+1)^2=5 and the vertex of the parabola f(x)= -3x^2+12x-10.
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Question 653877: Find the equation of the line satisfying the given conditions: Passes through the center of the circle (x+2)^2+(y+1)^2=5 and the vertex of the parabola f(x)= -3x^2+12x-10.
Answer by solver91311(24713) (Show Source): You can put this solution on YOUR website!
The equation of a circle centered at )
with radius 
is ^2\ +\ (y\ -\ k)^2\ =\ r^2)
The vertex of a parabola described as \ =\ ax^2\ +\ bx\ +\ c)
is at the point \right))
Step 1: Rewrite your circle equation:
\right)^2\ +\ \left(y\ -\ (-1)\right)^2\ =\ 5)
And then by inspection you can see that the first point for your line is )
Step 2: Calculate the 
-coordinate of your parabola vertex:
Step 3: Evaluate the parabola function at the coordinate of the vertex, i.e. calculate )
Step 4: Form an ordered pair from the results of steps 2 and 3.
Step 5: Use the results of steps 1 and 4 and the two-point form of an equation of a line to write your desired equation:
(x\ -\ x_1) )
where )
and )
are the coordinates of the given points.
John
My calculator said it, I believe it, that settles it
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