Respuesta :
Answer: the equation of the standard parabola
1) [tex](y-6)^2 = 4 (x-1)[/tex]
The equation of the standard parabola
2) (x+5)^2 = 16(y-2)
Step-by-step explanation:
Explanation
Parabola:-
The set of points in a plane whose distance from a fixed point and a constant ratio to their corresponding perpendicular distance from a fixed straight line is a conic.
Let S be a fixed point and l be a fixed straight line from any point P,the perpendicular PM is drawn to the line 'l'
- The locus of P such that [tex]\frac{SP}{PM} = constant[/tex]
- The fixed point 'S' is called the Focus.
- The fixed line'l 'is called the directrix of the conic
- The constant ratio is known as the eccentricity, denoted by 'e'
- If e=1 , the conic is called a parabola
1) Step 1 :-
Given the focus S = (2,6) and directrix is x=0
we know that [tex]\frac{SP}{PM}=1[/tex]
now cross multiplication , we get
[tex]SP = PM[/tex]
squaring on both sides,we get
[tex]SP^{2} = PM^2[/tex]
step 2:-
now using distance formula is
- [tex]\sqrt(((x_{2}-x_{1})^2+(y_{2} -y_{1} )^2)[/tex]
Given S =(2,6) and P(x,y) be any point on parabola
[tex]SP^2 = (x-2)^2+(y-6)^2[/tex]........(1)
Now using perpendicular distance formula
let P(x , y ) be any point on the parabola
- [tex]\frac{ax_{1}+by_{1}+c }{\sqrt{a^2+b^2} }[/tex]
Given the directrix is x =0 and P(x,y) be any point on parabola
[tex]PM^2 = \frac{x^2}{\sqrt{1}^2 }[/tex]......(2)
equating equation(1) and equation (2), on simplification
we get [tex](x-2)^2+(y-6)^2 = x^2[/tex].....(3)
- apply [tex](a-b)^2 = a^2+ b^2+2 ab[/tex]
now the equation (3) is
[tex](y-6)^2 = 4 x-4[/tex]
now the standard form of parabola is
[tex](y-k)^2 = 4 a(x-h)[/tex]
Final answer:-
[tex](y-6)^2 = 4 (x-1)[/tex]
2) Explanation:-
step 1:
Given vertex of a parabola is A(-5,2) and its focus is S(-5,6)
here the given points of 'x'co- ordinates are equal
- Therefore the axis AS is parallel to y- axis
now the standard equation of parabola
[tex](x-h)^2 = 4 a (y-k)[/tex]
now you have to find' a' value
Given vertex of a parabola is A(-5,2) and its focus is S(-5,6)
The distance of [tex]AS = \sqrt{(-5-(-5)^2+(2-6)^2}[/tex]
on simplification we get a =4
Final answer :-
the vertex (h,k) = (-5,2) and a=4
[tex](x-h)^2 = 4 a (y-k)[/tex]
The standard parabola is [tex](x+5)^2 = 16 (y-2)[/tex]
