# What is the intercepting point of the line y=5-x and the parabola y=(13-x^2)^1/2?

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### 3 Answers

To find the point of intersection of y = 5 - x and y=(13-x^2)^1/2, we just substitute y = 5 - x in y=(13-x^2)^1/2.

=> 5 - x = (13-x^2)^1/2

square both the sides

=> (5 - x)^2 = (13 - x^2)

=> 25 + x^2 - 10x = 13 - x^2

=> 2x^2 - 10x + 12 =0

=> 2x^2 - 6x - 4x + 12 = 0

=> 2x( x - 3) - 4 ( x - 3) = 0

=> (2x - 4)(x - 3) = 0

=> x = 2 and x = 3

For x = 2, y = 3 and for x = 3, y = 2.

**Therefore the points of intersection are (2,3) and (3,2)**

Given the line y= 5-x, and the parabola y= (13-x^2)^1/2

We know that the intersection points for the line and the parabola are x and y values such that y = y

==> 5-x = sqrt(13 - x^2)

We will square both sides:

==> (5-x)^2 = (13- x^2)

==> 25 - 10x + x^2 = 12 - x^2

We will combine like terms:

==> x^2 + x^2 - 10x + 25 - 12 = 0

==> 2x^2 - 10x + 13 = 0

Now we will solve for x using the formula:

x1= ( 10 + sqrt(100-4*13*2) / 2*2

= ( 10 + sqrt(4 ) /4 = (10+2)/4 = 12/3 = 3

==> x1= 3 ==> y1= 5-3 = 2

==> x2= ( 10-2)/4 = 8/4 = 2

==> x2= 2 ==> y2= 5-2 = 3

**Then, the intersection points are:**

**(2,3) and (3,2)**

To find the intercepting points, we'll have to solve the symmetric system.

We'll use the sum and the product to solve it.

We'll note x+y = S and xy = P.

We'll re-write the second equation as:

x^2 + y^2 = (x+y)^2 - 2xy

x^2 + y^2 = S^2 - 2P

We'll re-write the system in S and P.

S = 5

S^2 - 2P = 13

We'll susbtitute S in the second equation:

25 - 2P = 13

-2P = 13 - 25

-2P = -12

P = 6

Now, we'll form the quadratic knowing the sum and the product:

x^2 - 5x + 6 = 0

x1 = [5+sqrt(25 - 24)]/2

x1 = (5+1)/2

x1 = 3

x2 = (5-1)/2

x2 = 2

**So, the intercepting points are the solutions of the symmetric system and they are: {2 ; 3} or {3 ; 2}.**