Question

The curve 3x^2 + xy - y^2 +4y -3 =0 and the line y=2(1-x) intersect at the points A and B

Find the cords A and B

Answer 1

First, we need to find the coordinates of the intersection points between the curve and the line. We can substitute y = 2(1-x) into the equation of the curve and simplify to get:

3x^2 + x(2(1-x)) - (2(1-x))^2 + 4(2(1-x)) - 3 = 0

Simplifying this expression, we get:

3x^2 - 10x + 5 = 0

We can solve for x by using the quadratic formula:

x = (-b ± sqrt(b^2 - 4ac)) / 2a

where a = 3, b = -10, and c = 5. Substituting these values, we get:

x = (-(-10) ± sqrt((-10)^2 - 4(3)(5))) / 2(3)

x = (10 ± sqrt(40)) / 6

x = (5 ± sqrt(10)) / 3

Now we can substitute these values of x back into the equation y = 2(1-x) to find the corresponding values of y:

When x = (5 + sqrt(10)) / 3, y = 2(1 - (5 + sqrt(10)) / 3) = -(4/3) - (2/3)sqrt(10)

When x = (5 - sqrt(10)) / 3, y = 2(1 - (5 - sqrt(10)) / 3) = -(4/3) + (2/3)sqrt(10)

Therefore, the coordinates of point A are ((5 + sqrt(10)) / 3, -(4/3) - (2/3)sqrt(10)), and the coordinates of point B are ((5 - sqrt(10)) / 3, -(4/3) + (2/3)sqrt(10)).

Answer 2

`\textsf{A}=\left(-(1)/(3), (8)/(3)\right)`

`\textsf{B}=\left(1,0\right)`

Explanation:

Given equations:

`3x^2 + xy - y^2 +4y -3 =0`

`y=2(1-x)=2-2x`

To find the points at which the given curve and line intersect, substitute the second equation into the first equation to eliminate the y-variable:

`3x^2 + x(2-2x) - (2-2x)^2 +4(2-2x) -3 =0`

`3x^2+2x-2x^2-(4-8x+4x^2)+8-8x-3=0`

`3x^2+2x-2x^2-4+8x-4x^2+8-8x-3=0`

`3x^2-2x^2-4x^2+2x+8x-8x-4+8-3=0`

`-3x^2+2x+1=0`

Divide both sides of the equation by -1:

`3x^2-2x-1=0`

Rewrite -2x as (-3x + x):

`3x^2-3x+x-1=0`

Factor the first two terms and the last two terms separately:

`3x(x-1)+1(x-1)=0`

Factor out the common term (x - 1):

`(3x+1)(x-1)=0`

Apply the zero-product property:

`3x+1=0 \implies x=-(1)/(3)`

`x-1=0 \implies x=1`

To find the y-values of the points of intersection, substitute the found x-values into the original linear equation:

`\begin{aligned}x=-(1)/(3) \implies y&=2\left(1-\left(-(1)/(3)\right)\right)\\\\y&=2\left((4)/(3)\right)\\\\y&=(8)/(3)\end{aligned}`

`\begin{aligned}x=1 \implies y&=2\left(1-\left(1\right)\right)\\\\y&=2\left(0\right)\\\\y&=0\end{aligned}`

Therefore, the coordinates of the two points of intersection are:

`\textsf{A}=\left(-(1)/(3), (8)/(3)\right)`

`\textsf{B}=\left(1,0\right)`