Question

The
Curve
3x
$3x {?}^{2}$
+ xy- y² +4y-3=0 and the line
y = 2(1-x) intersect at
at the points A and B
Find the coordinates of A and of p​

Answer 1

Answer:

Equation of the curve = 3x² + xy - y² + 4y - 3 = 0   ...(1)

Equation of the line = y = 2 ( 1 - x ) ⇒ 2 - 2x   ... (2)

Substituting value of y from (2) in (1) we get:

⇒ 3x² + x ( 2 - 2x ) - ( 2 - 2x )² + 4 ( 2 - 2x ) - 3 = 0

⇒ 3x² + 2x - 2x² - [ 4 + 4x² - 8x ] + 8 - 8x - 3 = 0

⇒ 3x² - 2x² + 2x - 4 - 4x² + 8x + 8 - 8x - 3 = 0

⇒ 3x² - 2x² - 4x² + 8x - 8x + 2x + 8 - 4 - 3 = 0

⇒ - 3x² + 2x + 1 = 0

⇒ -3x² + 3x - x + 1 = 0

⇒ 3x ( - x + 1 ) + 1 ( - x + 1 ) = 0

⇒ ( 3x + 1 ) ( - x + 1 ) = 0

⇒ x = -1/3, 1

Hence if x is -1/3, then y would be:

⇒ y = 2 - 2 ( -1/3)

⇒ y = 2 + 2/3

⇒ y = 8/3

If x is 1, then y would be:

⇒ y = 2 - 2 ( 1 )

⇒ y = 0

Hence the two points are ( -1/3, 8/3 ) and ( 0,1 )

(Refer to the attachment for graph)

Answer 2

Step-by-step explanation:

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Given ,

3x²+xy-y²+4y-3=0 let it be eq- 1

y=2(1-x)

y=2-2x let it be eq- 2

Replacing y value in eq-1 ,

3x²+x(2-2x)-(2-2x)²+4(2-2x)-3=0

3x²+2x-2x²-(4+4x²-8x)+8-8x-3=0

3x²+2x-2x²-4-4x²+8x+8-8x-3=0

-3x²+2x+1=0

-3x²+3x-x+1=0

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

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

-x+1=0 & 3x+1=0

-x=-1 & 3x= -1

x= 1 & x= -⅓

Replacing any value of x in eq-2

If x=1

y=2-2x

y=2-2(1)

y=2-2

y=0

So, (x,y)= (1,0)

If x= -⅓

y=2-2x

y= 2-2(-⅓)

y= 2+⅔

y= 2×3+2÷ 3

y= 8/3

So, (x,y) = (-⅓ , 8/3)

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