Question

For what values of p does the vertex of the parabola y = x2 + 2px + 13 lie at a distance of 5 from the origin ?

Answer 1

Step-by-step explanation:

Given For what values of p does the vertex of the parabola y = x2 + 2px + 13 lie at a distance of 5 from the origin ?

• Let y = x^2 + 2px + 13
• Now add and subtract p^2
• So x^2 + 2px + p^2 – p^2 + 13
• So y = (x + p)^2 – p^2 + 13
• So (x + p)^2 = y + p^2 – 13
• Now general equation is (x – h)^2 = 4a(y – k)
• So (h,k) are the vertex of parabola.
• So coordinates of vertex will be (- p, 13 – p^2)
• So we have two points (0,0)
• Using distance formula we get
•                               5 = √(- p – 0)^2 + (13 – p^2 – 0)^2
•                                   Squaring we get
•                                   25 = p^2 + 169 + p^4 – 26 p^2
•                                  25 = 169 + p^4 – 25 p^2
•                            So p^4 – 25 p^2 + 144 = 0
•                         Or p^4 – 9p^2 – 16p^2 + 144 = 0
•                      Or p^2 (p^2 – 9) – 16 (p^2 – 9) = 0
•                              (p^2 – 9) (p^2 – 16) = 0
•                          So p^2 – 9 = 0          p^2 – 16 = 0
•                           Or p = ± 3          p = ± 4
• So for these values of p the vertex of the parabola is at a distance of 5 from the origin.

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https://brainly.in/question/6230209