Question
Question: What is the vertex form of an equation of the parabola whose focus is at \[\left( {200{\text{ }}, - ...
What is the vertex form of an equation of the parabola whose focus is at (200 ,−150) and a directrix of y=135?
Solution
Use the property that a point will be equidistant from the focus and from the directrix.
Use distance formula and equate it to other distance
Keep in mind that you need to remove square roots to solve.
Complete Step by step answer:
Let the point be (x,y)
Distance of two points ( x1,y1) and (x2,y2) formula is (x2−x1)2+(y2−y1)2
Distance between p(x,y) and focus (200,−150) is
D1 = (x−200)2+(y+150)2
D2 is the distance between the point (x,y) and a horizontal line y=p. The x-coordinates will be the same, so the distance between the point and line is the difference in the y-values.
D2 = modulus of y – (135)
We know that any point will be equidistant from focus and directrix
Hence equate D1=D2
(x−200)2+(y+150)2 = [ y – 135 ]
Squaring on both sides, we need to square on both sides to remove square roots
{\left( {x - 200} \right)^2} + {\left( {y + 150} \right)^2} = $$$${\left( {{\text{ }}y{\text{ }}-{\text{ }}135} \right)^2}
Expanding in( a−b)2= a2− 2ab + b2
→x2− 400 x + 40000 + y2+ 300y +22500 = y2− 270y + 18225
→x2− 400 x + 270y +300 y +22500 − 18225 = 0
→x2− 400 x +570y + 4275 = 0
Expand using formulae and cancel the same terms and rearrange to simplify .
Additional information:
For getting more attention or to attract the examiner you can draw parabola and you can show focus point and directrix in that which adds value to your answer.
Note:
Squaring on both sides should be done and need to apply algebraic formulae
Take care of signs while solving.
When you are transposing the terms put proper sign notations.