Question

What is the equation of the quadratic graph with a focus of (4,3) and a directrix of y=-6

Answer 1

Answer:

f(x) = (1/18)x^2 - (4/9)x - 65/18

Step-by-step explanation:

Since the focus (4,3) is higher up than is the directrix y = -6, the vertex is between y = -6 and y = 3, halfway between these values, to be exact:  y = -3/2.  Thus, the y-coordinate of the vertex is (-6 + 3/2), or -4.5.  The x-coordinate of the vertex is 4, which comes from that 4 in (4,3).  Thus, the vertex is (4, -4.5) or (4, -4 1/2).

The standard equation for a vertical parabola with vertex at (h,k) is:

4p(y-k) = (x-h)^2, where p is the distance between vertex and focus or between vertex and directrix.  In this case p = 4.5.  Thus, the equation of this quadratic is 4(4.5)(y - [-4.5]) = (x - 4)^2, or

18(y+4.5) = (x-4)^2, or y+4.5 = (1/18)(x-4)^2.

Let's expand this and rewrite the result as a quadratic equation in standard form y = ax^2 + bx + c:

y + 4.5 = (1/18)(x^2 - 8x + 16).  Multiplying all terms by 18 to remove both fractions, we get:

18y + 81 = x^2 - 8x + 16.  Rearranging these terms, we get:

18y = x^2 - 8x + 16 - 81, or

18y = x^2 - 8x - 65, or

y = f(x) = (1/18)x^2 - (4/9)x - 65/18.

Answer 2

Answer:

The correct option is B) $y=\frac{x^{2}}{6}-\frac{4x}{3}-\frac{11}{6}$

Step-by-step explanation:

We need to find out the quadratic graph with given focus $(4,-3)$ and a directrix $y=-6$

Using the distance formula, we find that the distance between (x,y) and the focus (4,-3) is $\sqrt{(x-4)^{2}+(y+3)^{2}}$ and the distance between (x,y) and the directrix y=-6, is $\sqrt{(y+6)^{2}}$

On the parabola, these distances are equal:

$\sqrt{(y+6)^{2}}=\sqrt{(x-4)^{2}+(y+3)^{2}}$

$(y+6)^{2}=(x-4)^{2}+(y+3)^{2}$

$y^{2}+36+12y=x^{2}+16-8x+y^{2}+9+6y$

Simplify the above

$36+12y-6y=x^{2}-8x+9+16$

$6y=x^{2}-8x+9+16-36$

$6y=x^{2}-8x-11$

Divide both the sides by 6,

$y=\frac{x^{2}}{6}-\frac{8x}{6}-\frac{11}{6}$

Simplified further,

$y=\frac{x^{2}}{6}-\frac{4x}{3}-\frac{11}{6}$

Therefore, the correct option is B) $y=\frac{x^{2}}{6}-\frac{4x}{3}-\frac{11}{6}$