Question

PLEASE HELP GUYS i am struggling so much, two questions

Answer 1

Answer: the equation of the standard parabola

1) $(y-6)^2 = 4 (x-1)$

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 $\frac{SP}{PM} = constant$
• 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 $\frac{SP}{PM}=1$

now cross multiplication , we get

$SP = PM$

squaring on both sides,we get

$SP^{2} = PM^2$

step 2:-

now using distance formula is

• $\sqrt(((x_{2}-x_{1})^2+(y_{2} -y_{1} )^2)$

Given S =(2,6) and P(x,y) be any point on parabola

$SP^2 = (x-2)^2+(y-6)^2$........(1)

Now using perpendicular distance formula

let P(x , y ) be any point on the parabola

• $\frac{ax_{1}+by_{1}+c }{\sqrt{a^2+b^2} }$

Given the directrix is x =0 and P(x,y) be any point on parabola

$PM^2 = \frac{x^2}{\sqrt{1}^2 }$......(2)

equating equation(1) and equation (2), on simplification

we get $(x-2)^2+(y-6)^2 = x^2$.....(3)

• apply $(a-b)^2 = a^2+ b^2+2 ab$

now the equation (3) is

$(y-6)^2 = 4 x-4$

now the standard form of parabola is

$(y-k)^2 = 4 a(x-h)$

Final answer:-

$(y-6)^2 = 4 (x-1)$

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

$(x-h)^2 = 4 a (y-k)$

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 $AS = \sqrt{(-5-(-5)^2+(2-6)^2}$

 on simplification we get a =4

Final answer :-

the vertex (h,k) = (-5,2) and a=4

$(x-h)^2 = 4 a (y-k)$

The standard parabola is $(x+5)^2 = 16 (y-2)$