Determine the equation of the conic section

  • #1
603
6

Homework Statement


Asymptotes at y=±[itex]\sqrt{10}[/itex] x /5

Foci: (±[itex]\sqrt{7}[/itex],0)


Homework Equations





The Attempt at a Solution



It's obviously a hyperbola. What I can't wrap my head around is why a^2 + b^2 ≠ c^2

c = sqrt 7 ; correct?

This yields
7=25+10
obviously NOT true. What am I missing here?
 
Last edited:

Answers and Replies

  • #2
34,671
6,384

Homework Statement


Asymptotes at y=±[itex]\sqrt{10}[/itex] x /5

Foci: (±[itex]\sqrt{7}[/itex],0)


Homework Equations





The Attempt at a Solution



It's obviously a hyperbola. What I can't wrap my head around is why a^2 + b^2 ≠ c^2
c = sqrt 7 ; correct?
No. c = √7.

You know that a2 + b2 = 7.

If one vertex is at (a, 0), use the slope of the asymptote to write b in terms of a, and then solve for a. You should find that a < c.

This yields
7=25+10
obviously NOT true. What am I missing here?
 
  • #3
603
6
Sorry I meant c^2 = 7.

Still. The asymptotes of a hyperbola that shoots of in the x direction is positive and negative a/b.

I don't get this at all.
 
  • #4
34,671
6,384
Sorry I meant c^2 = 7.

Still. The asymptotes of a hyperbola that shoots of in the x direction is positive and negative a/b.

I don't get this at all.
If the hyperbola's vertices are at (±a, 0), the asymptotes are given by y = ±(b/a)x. What are the coordinates of the point on the line y = (b/a)x at which x = a? You are given the slope of the asymptotes, and you also know that a2 + b2 = 7.

You have two equations that involve a and b, so you should be able to solve for these variables.
 

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