Question

How to solve a system of parabolic equations?
x²+2y²+4y = 4
x²-y²+2x-2y = 0

Answer 1

Step-by-step explanation:

Area of rectangle is 192 cm².

Step-by-step explanation:

Given :-

Length of rectangle is 16 cm.

Diagonal of rectangle is 20 cm.

To find :-

Area of rectangle.

Solution :-

First we will find breadth of rectangle because we do not have breadth of rectangle for area.

Let, rectangle be ABCD.

BC be length of rectangle.

AC be diagonal of rectangle.

And, AB be breadth of rectangle.

We know,

All angles of rectangle are of 90°.

So,

∆ABC is a right angle right. ∆ABC will right angled from B.

By Pythagoras theorem :

• Perpendicular² = Hypotenuse² - Base²

Perpendicular = AB

Hypotenuse = AC = 20 cm.

Base = BC = 16 cm.

Put all values in Pythagoras theorem :

⟶ (AB)² = (AC)² + (BC)²

⟶ (AB)² = (20)² + (16)²

⟶ (AB)² = 400 - 256

⟶ (AB)² = 144

⟶ AB = √144

⟶ AB = 12

Thus,

AB is 12 cm.

Answer 2

Step-by-step explanation:

This system represents an ellipse and a set of straight lines. If you solve each equation above for y, you can enter the "plus-minus" equations into your graphing calculator to verify this:

x2 – xy + y2 = 21

y2 – xy + (x2 – 21) = 0

y = [x ± sqrt(84 - 3x^2)]/2

x2 + 2xy – 8y2 = 0

0 = 8y2 – 2xy – x2

y = (x ± 3abs(x))/8

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As you can see, I used the Quadratic Formula in each of the original equations, in order to solve for y in terms of x. This gave me equations that I could graph. This technique doesn't come up that often, but it can be a life-saver when you can't seem to solve things any other way.

Oh, and where did those absolute-value bars come from? Recall that, technically, the square root of x2 is the absolute value of x. That's how I did that simplification in the next-to-last line above. And this absolute value will matter shortly.

The absolute value of x in the second equation above gives two cases for the values of y:

If x < 0, then | x | = –x, so y = (x ± 3x)/8 = x/2, – x/4

If x > 0, then | x | = x, so y = (x ± 3x)/8 = – x/4, x/2.

In either case, y = – x/4 or y = x/2.

Since I derived these "y=" solution-equations from the second of the original equations, I will plug them into the first equation to solve for some actual numerical values:

If y = – x/4: Copyright © 2002-2011 Elizabeth Stapel All Rights Reserved

x = ± 4

Then, plugging into the "y=" solution-equations above, I get:

x = -4 then y = 1; x = 4 then y = -1

If y = x/2:

x = ± 2sqrt(7)

Then:

x = -2sqrt(7) then y = -sqrt(7); x = 2sqrt(7) then y = sqrt(7)

Then the four solutions are:

(-2sqrt(7), -sqrt(7)), (2sqrt(7), sqrt(7)), (-4, 1), (4, -1)