A few math questions!?

Question

Please help me on the following (two parenthesis like this (+) is a plus):

1) Determine the nature of the roots without solving or graphing the equation

a) 5x^2-3x (+) 2=0

b) t^2 (+) t=1

2) Find the equation of the parabola with focus (2,3) and directrix y=-1

(I don't know what focus and directrix are!!!)

3) Find the equation of the ellipse with x-intercepts 6^(1/2), - (6^(1/2)) and y-intercepts 2,- 2

4) Graph:

x^2 (+) y^2 - 2x (+) 4y (+) 1 = 0 (I have a graphing calculator but i need to get it into y= form)

Thank you very much

Answer 1

ok

1) Nature of the roots is all about the discriminant b² - 4ac

if b² - 4ac < 0 (negative) you'll have no real roots. In other words you'll have imaginary roots

if b² - 4ac = 0 you'll have 1 real root or in other words 2 equal roots

if b² - 4ac > 0 (positive) you'll have 2 unequal real roots

so with that said...

a) 5x² - 3x + 2 = 0 here a = 5, b = -3, c = 2

find the discriminant b² - 4ac

(-3)² - 4(5)(2) = -39

since this is negative, you'll have imaginary roots

b) t² + t = 1

move the 1 over

t² + t - 1 = 0 here a = 1, b = 1, c = -1

(1)² - 4(1)(-1) = 5

this is positive, you'll have 2 unequal real roots. And since this is not a perfect square, they will be irrational as well.

2) standard form for a parabola...

(x - h)² = 4p(y - k)

(h, k) is the vertex

p is the directed distance from the vertex to the focus

in this problem, the focus is (2,3) and the directrix is the line, y = -1

note: the vertex is halfway in between the focus and the directrix. A parabola is the set of all points equidistant from a fixed point called the focus and a line, the directrix.

it helps to do a little sketch, plot the point (2,3) and draw the line y = -1. Looking at that the vertex has to be at the point (2,1). (2,1) is halfway in between the directrix and focus.

p is the distance from the vertex to the focus. p  = 2, so...

(x - h)² = 4p(y - k)

(x - 2)² = (4)(2)(y - 1)

(x - 2)² = 8(y - 1) you can leave it like this

note: there is tad more to parabolas. The standard from varies based on which way the parabola is opening.

3) standard form for an ellipse...

(x - h)² / a²  +  (y - k)² / b² = 1

(h, k) is the center and 2a is the length of the major axis and 2b is the length of the minor. You are given the intercepts

Points you know...

(√6, 0) and (-√6, 0) (x-intercepts)

(0,2) and (0,-2) (y-intercepts)

if you plot these points, you'll notice the center is at the origin (0,0)

also,  since √6 is greater than 2, this has a horizontal major axis. The length of the major axis is 2√6. Remember that's equal to 2a, which means a = √6 and a² = 6

the minor axis is equal to 4 and that's equal to 2b which means b = 2 so b² = 4  

that's all I need

(x - 0)² / 6  +  (y - 0)² / 4 = 1

clean it up

x²/6  +  y²/4 = 1

note: there is more to ellpises than I layed out. a² is not necessarily underneath the x². If this ellipse had a vertical major axis, a² would be under the y².

4) This is a circle. You do not want to get this in y =. You can but it's pointless. You want standard form.

group the terms and move the 1 over...

(x² - 2x) + (y² + 4y) = -1

complete the square...

(x² - 2x + 1) + (y² + 4y + 4) = -1 + 1 + 4

simplify and rewrite...

(x - 1)² + (y + 2)² = 4

center: (1, -2)

radius: 2

ok, so what did we learn??? There is a lot of material you need to study. The math itself is very easy, it's just a lot to memorize (standard forms, definitions, what letters mean etc.). But it's all very do-able. I did it, and I'm not that smart!

Answer 2

1) use b^-4ac. if the answer is larger than 0, there are 2 diff real roots. if it equals to 0, there are 2 equal real roots and if it is less than 0, there are no real roots.