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=0In 2,3,4 write the equation in...

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.