SOLUTION: Dear Sir/Ma'am, Please help me solve this problem Find the General Equation of the Parabola with Vertex at (5,3) and focus at (6,-3). Sketch and determine the parts of the Pa

Algebra ->  Quadratic-relations-and-conic-sections -> SOLUTION: Dear Sir/Ma'am, Please help me solve this problem Find the General Equation of the Parabola with Vertex at (5,3) and focus at (6,-3). Sketch and determine the parts of the Pa      Log On


   



Question 1181022: Dear Sir/Ma'am,
Please help me solve this problem
Find the General Equation of the Parabola with Vertex at (5,3) and focus at (6,-3). Sketch and determine the parts of the Parabola
Parts of the Parabola
1. Vertex
2. Focus
3. Directrix
4. Axis of Symmetry
5. E1, E1
6. Length of E1, E2
Thank you very much. More Power and God bless you.
Sincerely yours,
Lorna

Answer by MathLover1(20850) About Me  (Show Source):
You can put this solution on YOUR website!
the standard equation for an up-down facing parabola:
4p%28y-k%29=%28x-h%29%5E2
given:
Vertex at (5,3) =>h=5 and k=3
so far, equaion is
4p%28y-3%29=%28x-5%29%5E2
and given focus at (6,-3)=> I believe it is 3 not -3 because focus and verex must lie on same line
use it to calculate p
since F(h,k%2Bp)
if k%2Bp=-3=>3%2Bp=-3=>p=-6
and your equation is:
4%28-6%29%28y-3%29=%28x-5%29%5E2
-24%28y-3%29=%28x-5%29%5E2++

Sketch and determine the parts of the Parabola

Parts of the Parabola
1. Vertex is at (5,3)
2. Focus is at (5,-3)
3. Directrix: y=k-p=>y=3-%28-6%29=>y=3%2B6=>y=9
4. Axis of Symmetry: y=5
5. use (5,-3), E1, E1 will be where a line y=-3 intersect parabola
-24%28y-3%29=%28x-5%29%5E2++ ...plug in y=-3
-24%28-3-3%29=%28x-5%29%5E2++
144=%28x-5%29%5E2++
sqrt%28144%29=x-5+
12=x-5+
x+=+-7, when y+=+-3
x+=+17, when y+=+-3
E1=(-7,-3), E1=(17,-3)
6. Length of E1, E2 is distance between (-7,-3) and (17,-3), and it is 24