Question 770419
1.Find the equation of a hyperbola which has the foci of the ellipse 4x^2+9y^2=36 as vertices and the vertices of the ellipse as foci. sktch the graph.
<pre>
4x² + 9y² = 36

Divide through by 36

{{{4x^2/36}}}{{{""+""}}}{{{9y^2/36}}}{{{""=""}}}{{{1}}}

{{{x^2/9}}}{{{""+""}}}{{{y^2/4}}}{{{""=""}}}{{{1}}}

{{{(x-0)^2/3^2}}}{{{""+""}}}{{{(y-0)^2/2^2}}}{{{""=""}}}{{{1}}}

That is of the form

{{{(x-h)^2/a^2}}}{{{""+""}}}{{{(y-k)^2/b^2}}}{{{""=""}}}{{{1}}}

So the ellipse has center (0,0), a=3, so the vertices are (3,0),
b = 2, so covertices are (0,±2)

The distance from center to focis is c, and c² = a²-b², so we calculate c

c² = a²-b²
c² = 3²-2²
c² = 9-4
c² = 5
 c = &#8730;<span style="text-decoration: overline">5</span>

Here is that ellipse:

{{{drawing(400,400,-5,5,-5,5,graph(400,400,-5,5,-5,5),

arc(0,0,6,4),circle(-sqrt(5),0,.1),circle(sqrt(5),0,.1)      ) }}}


The hyperbola has equation of the form

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


center (0,0), vertices (±&#8730;<span style="text-decoration: overline">5</span>,0), and foci (±3,0). To find b, we use this equation:

c² a² + b²

3² = (±&#8730;<span style="text-decoration: overline">5</span>)² + b²
9 = 5 + b²
4 = b²
2 = b

So the hyperbola equation is

{{{(x-0)^2/(sqrt(5))^2}}}{{{""-""}}}{{{(y-0)^2/2^2}}}{{{""=""}}}{{{1}}}

{{{x^2/5}}}{{{""-""}}}{{{y^2/4}}}{{{""=""}}}{{{1}}}

Here is the hyperbola:

{{{drawing(400,400,-5,5,-5,5,graph(400,400,-5,5,-5,5,sqrt((4/5)x^2-4)),
graph(400,400,-5,5,-5,5,-sqrt((4/5)x^2-4)),

triangle(-sqrt(5),-2,0,-2,sqrt(5),-2),
triangle(-sqrt(5),2,0,2,sqrt(5),2),
triangle(-sqrt(5),-2,-sqrt(5),0,-sqrt(5),2),
triangle(sqrt(5),-2,sqrt(5),0,sqrt(5),2),

circle(3,0,.1), circle(-3,0,.1),
green(arc(0,0,6,4)),circle(-sqrt(5),0,.1),circle(sqrt(5),0,.1),
line(18,16.1,-18,-16.1),line(-18,16.1,18,-16.1)      ) }}}
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<pre>
2.Two members of the family of circles have radii all equal to 2 units, and are centred at (-2,2, and (2,-2)respectively.Find a member of this family that passes through the point (2,2). Sketch the graph.

{{{drawing(400,400,-6,6,-6,6,circle(-2,2,2),circle(2,-2,2), graph(400,400,-6,6,-6,6), circle(2,2,.1), graph(400,-6,6,-6,6)  )}}}   

If this means all the circles with radii =2, then there are an infinite number
of circles with radius = 2

Choose any center which is 2 units from the point (2,2), soch as
 (4,2), (2,4), (0,2), (2,0).  They are the four green circles below:

 {{{drawing(400,400,-6,6,-6,6,circle(-2,2,2),circle(2,-2,2), graph(400,400,-6,6,-6,6), circle(2,2,.1), graph(400,-6,6,-6,6),
green(circle(4,2,2),circle(2,4,2),circle(0,2,2),circle(2,0,2))

  )}}}

The one centered at  (4,2) has equation (x-4)²+(y-2)²=4
The one centered at  (2,4) has equation (x-2)²+(y-4)²=4
The one centered at  (0,2) has equation x²+(y-2)²=4
The one centered at  (2,0) has equation (x-2)²+y²=4

Edwin</pre>