Question

Prove that the equation of the parabola whose top and navel is located on the distance of the original point on the x-axis in sequence a and a is y square = 4 (a'-a) (x-a)

Answer 1

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$\bold{\underline{Question-}}$

Prove that the equation of the parabola whose top and navel is located on the distance of the original point on the x-axis in sequence a and a is y square = 4 (a'-a) (x-a)


$\bold{\underline{Answer-}}$

Suppose A is the top of parabola and S is navel, then

OA = a, OS = a'

•°• AS = OS - OA

=> a' - a

•°• The directors of Top A and Naval S are respectively (a, 0) and (a ', 0)

Assume AS is the axis of parallel whose equation is y = 0

Increase the SA to Z

Now,

SA = AZ => a' - a

OZ = OA - ZA

= a - (a' - a)

= 2a - a'


Draw line ZM through Z, which is perpendicular to the parabola, then the zm parabola gets its zodiac whose equation

x = 2a - a'


x - 2a + a' = 0


Suppose a point on parabola is P (x, y) ,


Define the determinant with P and add PS,
then by definition. SP = PM

means'


$\sqrt{ {(x - a')}^{2 } + (y - 0) {}^{2} } = x - 2a + a'$


${(x - a')}^{2} + {y}^{2} = (x - 2a + a') {}^{2}$


y² = (x - 2a + a')² - (x - a) ²


y² = 4(a' - a) (x - a)


Which is the desired equation of parabola










#@swarnimkumar22
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Answer 2

Answer:

Step-by-step explanation:

Suppose A is the top of parabola and S is navel, then

OA = a, OS = a'

•°• AS = OS - OA

=> a' - a

•°• The directors of Top A and Naval S are respectively (a, 0) and (a ', 0)

Assume AS is the axis of parallel whose equation is y = 0

Increase the SA to Z

Now,

SA = AZ => a' - a

OZ = OA - ZA

= a - (a' - a)

= 2a - a'

Draw line ZM through Z, which is perpendicular to the parabola, then the zm parabola gets its zodiac whose equation

x = 2a - a'

x - 2a + a' = 0

Suppose a point on parabola is P (x, y) ,

Define the determinant with P and add PS,

then by definition. SP = PM

means'

y² = (x - 2a + a')² - (x - a) ²

y² = 4(a' - a) (x - a)