What's the equation of the parabola that has its vertex at (8,–14) and a point (5,13) that lies on the curve? Question 13 options:

A)

y = (x + 8)2 – 14

B)

y = (x – 8)2 – 14

C)

y = 3(x + 8)2 – 14

D)

y = 3(x – 8)2 – 14

The expression with the described characteristics is: . Therefore the correct answer is letter D.

Step-by-step explanation:

I believe there is a typo in the vertex's coordinate, I think it should be (8,14). I came to this conclusion by observing the possible answers, if the given vertex is correct then there are no correct answers among them. I made my best effort to explain the question in a way that can be generalized for different vertex's coordinates.

The equations are given in their vertex form. This form has the following structure:

Where (h,k) are the vertex's coordinates, in our case (8, 14). This knowledge makes easier to determine a equation with the correct vertex as shown below:

Where the value of "a" must be determined by applying the given point.

The expression with the described characteristics is: . Therefore the correct answer is letter D.

< coa = 150 ( given)

as ba and bc are tangents

so oc and oa is perpendicular to cb and ab respectively because in circle normal passes through centre

so in quadrilateral oabc

< coa + < oab + < ocb + < abc = 360

150 +90+90 + < abc = 360

330 + < abc = 360

< abc = 360 -330 = 30

so < abc = 30

⭐dhruv⭐

answer: the distance

step-by-step explanation:

to calculate the distance d of a line segment with endpoints (x1, y1) and (x2, y2) use the formula d (x2 x1)2 (y2 y1)2. to calculate the midpoint of a line segment with endpoints (x1, y1) and (x2, y2) use the formula