Question 1142069
<pre>I'll only do one.  I'll do (v)

{{{4x^2-25y^2-8x-96=0}}}

Get terms in x² and x together.  Since there is no term in y, write y² as
(y-0)²  

{{{4x^2-8x-25(y-0)^2=96}}}

Factor the coefficient of x² out of the x² and the x terms.

{{{4(x^2-2x)-25(y-0)^2=96}}}

Add and subtract 1 inside the first parentheses:

{{{4(x^2^""-2x+1-1)-25(y-0)^2=96}}}

Factor the first three terms inside the first parentheses:

{{{4((x-1)^2^""-1)-25(y-0)^2=96}}}

Distribute the 4 into the big parentheses, removing the big parentheses,
leaving the smaller parentheses intact:

{{{4(x-1)^2-4-25(y-0)^2=96}}}

Add 4 to both sides

{{{4(x-1)^2-25(y-0)^2=100}}}

Get 1 on the right side by dividing every term by 100

{{{4(x-1)^2/100^""-25(y-0)^2/100^""=100^""/100^""}}}

Simplify to get in standard form:

{{{(x-1)^2/25^""-(y-0)^2/4^""=1}}}

Since x comes first, the hyperbola opens left and right.

Compare to equation  {{{(x-h)^2/a^2-(y-k)^2/b^2=1}}}

a² = 25, so a = 5;   b² = 4, so b = 2

Center = (h,k) = (1,0), semi-transverse axis = a = 5,
semi-conjugate axis = b = 2    

Draw defining rectangle with center (h,k) = (1,0) with horizontal dimension
equal to the entire transverse axis, or 2a = 2(5) = 10 units and vertical
dimension equal to the the entire conjugate axis, or 2b = 2(2) = 4.

{{{drawing(400,200,-9,11,-5,5,graph(400,200,-9,11,-5,5),

green(line(-4,-2,-4,2),line(-4,-2,6,-2),line(-4,2,6,2),line(6,2,6,-2)) )}}}

Draw and extend the diagonals of the defining rectangle.  They are the
asymptotes.

{{{drawing(400,200,-9,11,-5,5,graph(400,200,-9,11,-5,5),

green(line(-4,-2,-4,2),line(-4,-2,6,-2),line(-4,2,6,2),line(6,2,6,-2),line(-14,6,16,-6),line(-14,6,16,-6),line(-14,-6,16,6),line(-14,-6,16,6)     ) )}}}

Now we can sketch in the hyperbola.

{{{drawing(400,200,-9,11,-5,5,graph(400,200,-9,11,-5,5,(2/5)sqrt(x^2-2x-24)),graph(400,200,-9,11,-5,5,(-2/5)sqrt(x^2-2x-24)),

green(line(-4,-2,-4,2),line(-4,-2,6,-2),line(-4,2,6,2),line(6,2,6,-2),line(-14,6,16,-6),line(-14,6,16,-6),line(-14,-6,16,6),line(-14,-6,16,6)     ) )}}}

The vertices are the midpoints of the left and right sides of the defining
rectangle, (-4,0) and (6,0).

We calculate the foci, by using the Pythagorean relationship for all
hyperbolas, which is c²=a²+b², where c is the distance from the center to
the foci.

{{{c^2=a^2+b^2}}}
{{{c^2=5^2+2^2}}}
{{{c^2=25+4}}}
{{{c^2=29}}}
{{{c="" +- sqrt(29)}}}
{{{c="" +- 5.4}}}, approximately

The word "foci" is the plural of the word "focus".

The left focus is c units left of the center (1,0) or {{{(matrix(1,3,1-sqrt(29),",",0))}}}


The right focus is c units right of the center (1,0) or {{{(matrix(1,3,1+sqrt(29),",",0))}}}

They are the two black dots drawn below:

{{{drawing(400,200,-9,11,-5,5,graph(400,200,-9,11,-5,5,(-2/5)sqrt(x^2-2x-24)),graph(400,200,-9,11,-5,5,(2/5)sqrt(x^2-2x-24)),
circle(1-sqrt(29),0,.15),
circle(1-sqrt(29),0,.13),
circle(1-sqrt(29),0,.11),
circle(1-sqrt(29),0,.09),
circle(1+sqrt(29),0,.15),
circle(1+sqrt(29),0,.13),
circle(1+sqrt(29),0,.11),
circle(1+sqrt(29),0,.09),



green(line(-4,-2,-4,2),line(-4,-2,6,-2),line(-4,2,6,2),line(6,2,6,-2),line(-14,6,16,-6),line(-14,6,16,-6),line(-14,-6,16,6),line(-14,-6,16,6)     ) )}}}

To find the equations of the asymptotes, they are the lines that go:

(1) through the center (1,0) and the upper right corner of the defining
rectangle (6,2), which has slope 2/5, which is b/a, so the equation is

{{{y-y[1]=m(x-x[1])}}}
{{{y-0=expr(2/5)(x-1)}}}
{{{y=expr(2/5)(x-1)}}}

(2) through the center (1,0) and the lower right corner of the defining
rectangle (6,-2), which has slope -2/5, which is -b/a, so the equation is

{{{y-y[1]=m(x-x[1])}}}
{{{y-0=-expr(2/5)(x-1)}}}
{{{y=-expr(2/5)(x-1)}}}

Edwin</pre>