Question

For what values of p does the vertex of the parabola y=x^2 +2px + 13 lie at a distance of 5 from the origin?

Answer 1

The vertex of a parabola is the point where the parabola crosses its axis. If the coefficient of the x2 term is positive, the vertex will be the lowest point on the graph, the point at the bottom of the “U”-shape. If the coefficient of the x2 term is negative, the vertex will be the highest point on the graph, the point at the top of the “U”-shape.

The standard equation of a parabola is
y = ax^2 + bx + c.

Example:

Find the vertex of the parabola.

y = 3x^2 + 12x – 12

Here, a = 3 and b = 12. So, the x-coordinate of the vertex is:

-12/2(3)=-2

Substituting in the original equation to get the y-coordinate, we get:

y = 3(–2)^2 + 12(–2) – 12

= –24

So, the vertex of the parabola is at (–2, –24).

Answer 2

since the x-coordinate of the vertex for the equation y=ax^2+bx+c is given by -b/(2a) all you need to do is solve the equation:

\(\large |-\frac{b}{2a}|=5 \) where a=1, b=2p

Answer 3

is it okay or there's more need to explain

Answer 4

@dpaInc i want full solution.

Answer 5

me too...

Answer 6

@dpaInc -b/2a is distance from y axis. its not distance from origin..

Answer 7

so somewhere we have to use the distance formula of coordinates.

Answer 8

i got the vertex of parabola = (-p,-p^2 +13)
Origin =(0,0)
distance between both = 5

Answer 9

anyone check the vertex and solve ahead.

Answer 10

yup your vertex is correct. Now find the distance from the vertex to origin (in terms of p) and then equate it to 5.

Answer 11

still i am not getting the sol.
so pls give me the solution ahead.

Answer 12

distance of vertex from origin = \(\sqrt{(-p-0)^{2} + (-p^{2}+13 - 0)^{2}} = 5\)
squaring both sides, and simplifying, we get
\(p^{4} - 25p^{2}+144 = 0\) can you solve this further ?

Answer 13

then do this...
\(\large 5=\pm \sqrt{(\frac{-b}{2a})^2+(f(-\frac{b}{2a}))^2} \)

Answer 14

then..........

Answer 15

if you are still stuck ...
p^4-25p^2+144=0
let p^2 = t

t^2-25t+144=0
t^2-16t-9t+144 = 0
t(t-16) -9(t-16) = 0
(t-16)(t-9) = 0

(p^2-16)(p^2-9) = 0

(p+4)(p-4)(p+3)(p-3) = 0

p = +-(3, 4)