Question

Are the two parabolas defined below similar or congruent or both? Justify your reasoning. Parabola 1: The parabola with a focus of (0,2) and a directrix line of y= −4. Parabola 2: The parabola that is the graph of the equation y=(1/6)x².

Answer 1

Answer:

They are similar by a factor of 14

Step-by-step explanation:

From the general equation of a parabola opening up (since the directrix is on the y axis and the focus is upwards this line) the focus coordinates are:

$Focus = (h,k+p) = (0,2)$

So h=0 and k+p=2, for the directrix:

$y = k-p=-4$

solving k and p from the two equations above:

$k=-1 and p=3$

So for Parabola 1 the equation would be:

$x^{2} =4*3(y+1)$

For Parabola 2 the general equation is:

$x^{2} =4*(3/2)(y)$

here h=0, k=0, p=3/2 so focus is (0,3/2), directix is y=-3/2, they have the same orientation.

they are not congruent. For them to be similar it must comply:

$x_{1}=k x_{2}\\y_{1}=ky_{2} => (1/12)x_{1} ^{2} -1 = k(1/6)x_{2} ^{2}$

replacing $x_{1}$ and solving for k:

$(1/12)(kx_{2}) ^{2} -1 = k(1/6)x_{2} ^{2}\\ (1/12)k^{2} x_{2}^{2}-1 = k(1/6)x_{2} ^{2}\\k-12 = (12/6)=2\\k = 2+12=14$

They are similar by a factor of 14