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solve for x: sec^2x - tanx-3 = 0 2.    2x^2-18=9y^2 3.     x^2+y^2-6x+4y=3...

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donovanp | Student, Undergraduate | (Level 2) eNoter

Posted July 18, 2010 at 11:25 PM via web

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solve for x: sec^2x - tanx-3 = 0

2.    2x^2-18=9y^2

3.     x^2+y^2-6x+4y=3

4.     25y^2-225+9x^2=0

In 2,3,4 write the equation in standard form to identify the curve.

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neela | High School Teacher | (Level 3) Valedictorian

Posted July 19, 2010 at 12:23 AM (Answer #1)

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To solve for x:

1) sec^2-tanx -3 = 0

Solution:

sec^2x - tanx-3 = 0

(tan^2x+1)-tanx-3 = 0, as sec^2x =tan^2x+1 is an identity.

t^2-t-2 = 0, where t = tanx

(t-2)(t+1) = 0

t=2 , t=-1,  x = arc tan2 = 63.435 deg approx , or 180 +63.435 deg.

t=1: x = arc tan 1 = 45 deg or 180+45 deg.

 

2.    2x^2-18=9y^2

Solution:

2x^2-18 = 9y^2

2x^2 = 9y^2+18

x ^2 = (9y^2+18)/2 = (9/2)(x^2+9)

x = + Or-  {(9/2)(x^2-9)}^(1/2)

Standard form of 2x^2 -18 = 9y^2:

2x^2 -9y^2 = 18  equivalent form

2x^2/18 -9y^2/18 = 1 equivalent form

x^2/3^2 - y^2/(sqrt2)^2 = 1 is the standard form of hyperbola:

X^2/a^2-Y^2/b^2= 1

 

3.     x^2+y^2-6x+4y=3

Solution:

We  use the formula of solution for the quadratic equation ax^2+bx+c =0. This has solutions   {-b+or -sqrt(b^2-4ac)}/(2a)

Wrting as a quadratic equation in x, we get:

x^2 -6x+(y^2+4y-3) = 0

a=1, b =-6 and c = y^2+4a-3

x = {- -6 +or- sqrt( (-6)^2-4*1(y^2+4y-3))}/(2*1)

= { (6+or-sqrt(48-y^2-16y)}/2

= {3 +or-  sqrt(12-y^2-4)}

The standard form is the circle  x^2+y^2+2gx+2fy+c =0 with (-g,-f) as centre and (g^2+f^2-c )^(1/2) as radius.

Here x^2+y^2-6x+4y-3  is the circle  with (3, -2) as centre and radius , ((3)^2+(-2)^2-(-3))^(1/2) =4.

4.     25y^2-225+9x^2=0

Solution:

Re arranging, 9x^2 = 225-25y^2

(3x)^2 = 25(9-y^2), Take the sqre root.

3x = +or-   sqrt{25(9-y^2)}

3x = +or- 5(9-x^2)^(1/2)

Standard form of ellipse is X^2+Y^2+b^2 =1

25y^2-225+9y^2 = 0 could be written as :

25x^2+9y^2 = 225

25x^2/225 +9y^2/225 = 225/225

x^2/3^2+y^2/(5^2) =2,  , where 3 and 5 are the x semi minor  and y  (semi major) axis.

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sciencesolve | Teacher | (Level 3) Educator Emeritus

Posted April 5, 2012 at 2:30 AM (Answer #2)

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1) You need to use the following formulas for `sec^2` x such that:

`sec^2x = 1/(cos^2x)`

Hence, substituting `1/(cos^2x)`  for `sec^2x`  yields:

`1/(cos^2x) -tan x- 3 = 0`

You need to remember that `1/(cos^2x) = 1 + tan^2 x` , hence you need to substitute 1 + tan^2 x for 1/(cos^2x) in equation such that:

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

Collecting like terms yields:

`tan^2 x - tan x - 2 = 0`

You should come up with the substitution tan x = y such that:

`y^2 - y - 2 = 0`

You need to use quadratic formula such that:

`y_(1,2) = (1 +- sqrt(1 + 8))/2`

`y_(1,2) = (1 +- sqrt9)/2`

`y_(1,2) = (1 +- 3)/2 =gt y_1 = 2`

`y_2 = -1`

You should solve for x the equations tan x = 2 and tan x = -1 such that:

`tan x = 2 =gt x = tan^(-1) 2 + npi`

`tan x = -1 =gt x = -tan^(-1)(1) + npi`

`x = -pi/4 + npi`

Hence, evaluatin solutions to equation yields `x = tan^(-1) 2 + npi ` and `x = -pi/4 + npi`

`.`

2) You need to move the terms containing x and y to the left side such that:

`2x^2 - 9y^2 = 18`

You need to divide by 18 both sides such that:

`x^2/9 - y^2/2 = 1`

This equation represents the standard form of equation of hyperbola with a horizontal transverse axis.

3) The form of equation `x^2+y^2-6x+4y=3`  is similar to general form of equation of circle, hence you should collect the terms in x and the terms in y and you should try to complete the squares such that:

`(x^2-6x+9)+(y^2+4y+4)=3+9+4`

`(x-3)^2+(y+2)^2=16`

Hence, the equation represents the equation of a circle with center (3,-2) and radius r=4.

4) You need to isolate the terms in x and y to the left side such that:

`25y^2 + 9x^2 = 225`

You should divide by 225 such that:

`y^2/9 + x^2/25 = 1`

This equation represents the standard form of equation of ellipse with horizontal major axis.

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