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in the coordinate plane, we plot the point (4t-4,-2t+7) for every real number t to obtain a graph g. for example, when t=2, we have 4t-4=4 and -2t+7=3, so the point (4,3) lies on the graph g

(a) show that every point on the graph g lies on the line y=-x/2+5

(b) show that every point on the line y=-x/2+5 lies on the graph g

(a) In order to solve this problem we will introduce new variables u and v, where u = 4t - 4 and v = -2t + 7.  Notice that as t gets very large, u = 4t - 4 looks like 4t, and v = -2t + 7 looks like -2t.  We can formalize this by saying that as t gets very large, v/u = (-2t + 7)/(4t - 4).  Divide both sides by t, to get v/u = (-2 + 7/t)/(4 - 4/t).  As t gets large, 7/t and 4/t are small, so u/v becomes (-2 + 7/t)/(4 - 4/t) = -1/2.  This tells us that we should look for a line of slope -1/2.

If y/x = (-2t + 7)/(4t - 4), then y(4t - 4) = x(-2t + 7), so 4yt - 4y = -2xt + 7x.  Then 4y + 7x = 4yt + 2xt = t(4y + 2x).  Since this is true for every value of t, we must have 4y + 7x = 0 and 4y + 2x = 0.  When we solve this system, we get that x = y = 0, so that tells us that our line passes through the origin.  And we know that the slope of the line is -1/2, so the equation of our line is y = -x/2.  But that is only the base line - we can take our line and shift it in the x-direction, or y-direction, until it matches what we want.  Here, the problem tells us that our line passes through the point (4,3), so we can shift it 3 units up, to get y = -x/2 + 3.  We can then shift it four units in the x-direction, to get y = -(x + 4)/2 + 3 = -x/2 + 5.  So this is the equation of the line.

(b) We can use the same argument as in part (a).  Let u = 4t - 4 and v = -2t + 7.  As we said in part (a), as t gets very large, u/v gets close to -1/2.  So we can find a value of t so that u/v is close to -1/2.  If y = -x/2 + 5, then y/x is close to -1/2.  So choose the value of t so that y/x = u/v.  Then y/x = (-2t + 7)/(4t - 4).  This leads to the same equation 4y + 7x = 4yt + 2xt = t(4y + 2x).  Then again, 4y + 7x = 0 and 4y + 2x = 0, so x = y = 0.  This gives us a base line of y/x = -1/2, or y = -x/2.  We can then shift five units up, because we want the line to pass through (0,5), to get the line y = -x/2 + 5.  So the line of y = -x/2 + 5 is the graph of G.

the transverse axis of a hyperbola is the line that contains the two vertices and the two focuses.

in this example (hyperbola of equation x^2/2 - y^2/4 = 1), the transverse axis is the  x-axis  . 

step-by-step explanation:

jml angle lkj & angle mj = lm