Question

PLEASE HELP!!! I'LL GIVE YOU BRAINLIEST!

Answer 1

(x - a)^2 + (y - b)^2 = (y - d)^2 

x^2 - 2ax + a^2 + y^2 - 2by + b^2 = y^2 - 2dy + d^2 

Subtract y^2 from both sides... 

x^2 - 2ax + a^2 - 2by + b^2 = -2dy + d^2 

Now I would just substitute in the the values (a = -4, b = 17/8, d = 15/8)... 

x^2 - 2(-4)x + (-4)^2 - 2(17/8)y + (17/8)^2 = -2(15/8)y + (15/8)^2 

Simplify... 

x^2 + 8x + 16 - (17/4)y + 289/64 = -(15/4)y + 225/64 

Add (17/4)y to both sides... 

x^2 + 8x + 16 + 289/64 = (2/4)y + 225/64 

x^2 + 8x + 16 + 289/64 = (1/2)y + 225/64 

Subtract 225/64 from both sides... 

x^2 + 8x + 16 + 1 = (1/2)y 

x^2 + 8x + 17 = (1/2)y 

Multiply both sides by 2... 

2x^2 + 16x + 34 = y 

...which is equal to equation D.

Answer 2

Answer:

Option D - $f(x)=2x^2+34+16x$

Step-by-step explanation:

Given :  Focus $F=(-4,\frac{17}{8})$ and a directex  $y=\frac{15}{8}$

To find : What is the equation of the quadratic graph with a focus and directrix?

Solution :

The general form of the quadratic equation is  $(x-h)^2=4p(y-k)$

Where, Focus of the equation is  $F=(h,k+p)$ and

Directrix is  $y=k-p$    

Comparing the given focus and directrix with the general one.

$(-4,\frac{17}{8})=(h,k+p)$

h=-4 and

$k+p=\frac{17}{8}$ ....(1)

$k-p=\frac{15}{8}$ .....(2)

Add equation (1) and (2)

$2k=4$

$k=2$

Substitute in equation (1)

$2+p=\frac{17}{8}$

$p=\frac{17}{8}-2$

$p=\frac{17-16}{8}$

$p=\frac{1}{8}$

Now, We get h=-4 , k=2 and $p=\frac{1}{8}$

Substitute in the general form,

$(x-(-4))^2=4(\frac{1}{8})(y-2)$

$x^2+16+8x=\frac{1}{2}(y-2)$

$2x^2+32+16x=y-2$

$2x^2+32+16x=y-2$

i.e, The equation of the quadratic graph is $f(x)=2x^2+34+16x$

Therefore, Option D is correct.