Question

find the distance between p(x,y) from it origin

Answer 1

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Answer 2

The distance from (x,y) to (2,5) is given by distance formula:

d = √[(x-2)2+(y-5)2]

 

The distance from (x,y) to the line x=-1 will taken from

the perpendicular distance as shortest which will be a

horizontal distance since x=-1 is a vertical line.

This distance will be the difference in the x coordinates

stated as a positive or... |x-(-1)| or |x+1|

 

Since we want the distance from (x,y) to (2,5) to be

twice the distance from (x,y) to x=-1 we have...

 

√[(x-2)2+(y-5)2] = 2|x+1|

 

square both sides... Note that since squaring (x+1) will

automatically result in a positive we will no longer need the

absolute value sign

 

(x-2)2+(y-5)2 = 4(x+1)2

 

x2-4x+4+y2-10y+25 = 4(x2+2x+1)

x2-4x +y2-10y+29 = 4x2+8x+4

 

Get everything to one side of the = sign except the constant.

I will subtract the left from the right just because

I prefer my x2 to be a positive coefficient....

 

3x2 + 12x - y2 + 10y = 25

 

Note this is becoming the equation of a hyperbola.

We want standard form (x-h)2/a2 - (y-k)2/b2 = 1 :

We need to do some complete the square work here.

 

3(x2+4x) - (y2-10y) = 25

3(x2+4x+4) - (y2-10y + 25) = 25 + 12 - 25

 

3(x+2)2 - (y-5)2 = 12

 

(x+2)2/4 - (y-5)2/12 = 1

 


We have a hyperbola centered at (h,k) or (-2,5)

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