Solve Conics Hyperbola Homework Statement

  • Thread starter Thread starter Mentallic
  • Start date Start date
  • Tags Tags
    Conics Hyperbola
Click For Summary
SUMMARY

The discussion focuses on solving a hyperbola-related homework problem involving the calculation of distances and intersections. The participant successfully solved part (i) but sought a more efficient method for part (ii). They proposed using the distance formula and the Pythagorean Theorem but later discovered the formula for the perpendicular distance from a point to a line, given by d = |ax₁ + by₁ + c| / √(a² + b²), which simplifies the solution process significantly.

PREREQUISITES
  • Understanding of conic sections, specifically hyperbolas.
  • Familiarity with coordinate geometry concepts such as distance formulas.
  • Knowledge of linear equations and their properties, including perpendicular lines.
  • Basic proficiency in applying the Pythagorean Theorem in geometric contexts.
NEXT STEPS
  • Study the derivation and applications of the perpendicular distance formula from a point to a line.
  • Explore advanced techniques for solving conic section problems, particularly hyperbolas.
  • Practice problems involving simultaneous equations and their geometric interpretations.
  • Review the properties of perpendicular lines in coordinate geometry.
USEFUL FOR

This discussion is beneficial for students studying conic sections, particularly those tackling hyperbola problems in geometry. It is also useful for educators looking for efficient methods to teach distance calculations in coordinate geometry.

Mentallic
Homework Helper
Messages
3,802
Reaction score
95

Homework Statement


http://img11.imageshack.us/img11/6340/conicshyperbola1.jpg

Homework Equations


[tex]d^2=(x_2-x_1)^2+(y_2-y_1)^2[/tex]

[tex]y-y_1=m(x-x_1)[/tex]

[tex]m_1m_2=-1[/tex]


The Attempt at a Solution


I was able to answer (i) but for (ii) I would go about it like this:

Find the equation of the line SR by using that it is perpendicular to the line l and passing through the focus S(ae,0). Then solve both equations simultaneously to find the point of intersection at R(x,y). Then find the distance between S and R given that I know both coordinates.
But looking at the marking criteria, it is only worth 1 mark and thus must have a much simpler way of being solved. Any ideas?
 
Last edited by a moderator:
Physics news on Phys.org
The only other thing I can see is to let point [itex]Q[/itex] be the x-intercept of the line [itex]l[/itex] and let [itex]R=(x,y)[/itex]. Then you can apply the Pythagorean Theorem to triangle [itex]QRS[/itex] and solve for [itex](x,y)[/itex]. The nice thing about this is that both points [itex]Q[/itex] and [itex]S[/itex] have only one nonzero coordinate.
 
Well then [tex]Q(acos\theta,0)[/tex] but the problem with that technique is that I'll still need to use the distance formula after finding point R, which is basically not any faster than what my original thinking was.

I asked my teacher on this one, there is a basic formula that I completely forgot existed.

[tex]d=\frac{|ax_1+by_1+c|}{\sqrt{a^2+b^2}}[/tex]

gives the perpendicular distance from a point to a line.
 

Similar threads

Replies
2
Views
1K
  • · Replies 3 ·
Replies
3
Views
5K
  • · Replies 6 ·
Replies
6
Views
2K
Replies
4
Views
4K
  • · Replies 5 ·
Replies
5
Views
2K
Replies
1
Views
2K
  • · Replies 5 ·
Replies
5
Views
2K
Replies
4
Views
2K
Replies
4
Views
2K
  • · Replies 3 ·
Replies
3
Views
3K