Conics question: Ex3.3 Q7 from Cambridge (1 Viewer)

[McSo]

P(asec@, btan@) lies on the hyperbola x^2/a^2 -y^2/b^2 = 1. The tangent at P meets the asymptotes at the point M and N. Show that PM = PN.

I basically tried getting the points M and N and using the distance formula.. but it ended up looking very ugly and inefficient.

 

[Riviet]

You've got the right method, check your tangent and coordinates of intersection, M and N. I know the algebra gets ugly, but that's the nature of the topic.

 

[hyparzero]

The easier way would be to prove that P(asec@, btan@) is the midpoint of MN.

 

[McSo]

Yeah, a friend mentioned that to me, but they said it didn't work out too well... I havn't tried it though.. I might try it now, but anyway, if anyone can think of a better way (with the solution) please go ahead and reply. My teacher also said try similar triangles...

 

[hyparzero]

x²/a² - y²/b² = 1

Tangent Eqn => xx'/a² - yy'/b² = 1 ......... (sub P(asec@,btan@)

=> ab²xsec@ - ba²ytan@ = a²b² ... (1)

Asymptotes are +- bx/a .... (sub y = bx/a)
Hence (1) = > ab²xsec@ - ab²xtan@ = a²b²
=> ab²x[sec@ - tan@] = a²b²
=> x = a/(sec@ - tan@)

Similarly...... if y = -bx/a, then intersection x = a/(sec@ + tan@)

Hence you have the co-ordinates:
M[ a/(sec@ - tan@) , b/(sec@ - tan@)] and N[ a/(sec@ + tan@) , b/(sec@ + tan@)]

Use the midpoint formula, and simplify a bit, and everything should cancel out, and you'll get (asec@, btan@) which is P.

Hence PM = PN

 

[Riviet]

I've just done it hyparzero's way and it works out good, the algebra's not hard, it's just long.

EDIT: the equation of the tangent to the hyperbola at (asec@, btan@) is
xsec@ - ytan@ = 1
...a............b

 

[McSo]

Isn't the equation of a tangent for a hyperbola

xsec@/a - ysec@/b = 1.. ?

But other then that, I think I get it.. I'll try it now, thanks.

 

[hyparzero]

Isn't the equation of a tangent for a hyperbola

xsec@/a - ysec@/b = 1.. ?

But other then that, I think I get it.. I'll try it now, thanks.

if you implicity differentiate your hyperbola equation, you should see that a² and b² are constants.

 

[McSo]

I just did it then and got it, thanks for the help guys.