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3 years ago

What is the solution of 3x+y=2 & 4y=12-12x?

Mathematics

1 answer:

lorasvet [3.4K]3 years ago

5 0

Answer:

Step-by-step explanation:

3x+y=2............(1)

4y=12-12x........(2)

Making y as subject in (1) and substitute in (2)

y = 2 - 3x......(3)

4(2 - 3x) = 12-12x

8 - 12x = 12 - 12x

12x - 12x = 12- 8

0 = 4

So no solution since variable cancelled out

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Let point P be with coordinates $(x_0,y_0).$ Find the equation of the tangent line.

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2. The equation of the tangent line at point P is

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Find x-intercept and y-intercept of this line:

when x=0, then $y=y_0+2x_0^2;$
when y=0, then $x=\dfrac{y_0}{2x_0}+x_0=\dfrac{y_0+2x_0^2}{2x_0}.$

The area of the triangle enclosed by the tangent line at P, the x-axis, and y-axis is

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Since point P is on the parabola, then $y_0=1-x_0^2$ and

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Find the derivative A':

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Equate this derivative to 0, then

$12x_0^4+8x_0^2-4=0,\\ \\3x_0^4+2x_0^2-1=0,\\ \\D=2^2-4\cdot 3\cdot (-1)=16,\ \sqrt{D}=4,\\ \\x_0^2_{1,2}=\dfrac{-2\pm4}{6}=-1,\dfrac{1}{3},\\ \\x_0^2=\dfrac{1}{3}\Rightarrow x_0_{1,2}=\pm\dfrac{1}{\sqrt{3}}.$

And

$y_0=1-\left(\pm\dfrac{1}{\sqrt{3}}\right)^2=\dfrac{2}{3}.$

Answer: two points: $P_1\left(-\dfrac{1}{\sqrt{3}},\dfrac{2}{3}\right), P_2\left(\dfrac{1}{\sqrt{3}},\dfrac{2}{3}\right).$

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