Q. conics (1 Viewer)

[marsenal]

This is from Arnold,

P(asec@, btan@) lies on the hyperbola x²/a² - y²/b² =1. The tangent at P cuts the x-axis at X and the y-axis at Y. Show that PX/PY=sin²@ and deduce that if P is an extremity of a latus rectum, then PX/PY= (e²-1)/e²

I can get the first part but I don't understand how to get the second bit.

And also with locus questions, what kind can we be expected to get? Just ones to do with rectangular hyperbola or can we get ones with ellipses and regular hyperbolas and mixes as well?

 

[freaking_out]

Originally posted by marsenal
...And also with locus questions, what kind can we be expected to get? Just ones to do with rectangular hyperbola or can we get ones with ellipses and regular hyperbolas and mixes as well?

only rectangular hyperbolas.....it even says in ze sillabus.

 

[Fosweb]

The Latus rectum runs parallel to the directrix, and it passes through (ae,0), so instead of using x = asec@ (as your x coord) to find the distances PX and PY, use x = ae.

Try that. But I am stuck on the first part. I ended up with a dirty big load of trig stuff, which I can't seem to simplify down to sin^2(@)

What I did: find distances PX^2 and PY^2, and then divide that. Then I was hoping that would end up at sin^4(@) so that root of that was sin^2(@), but its not looking too nice.
Is this how you did this?

I have so far:

^PX² / _PY² = sin²@ [ ^{(a² + b²cos²@)(sin²@)} / _{(a²sin²@ + b²)(cos²@)} ]

Does the part in the [ ] brackets equal sin²@
Is this how you got the first bit marsenal?

 

[Constip8edSkunk]

second part:
ae = asec@ -> sec@=e -> cos@=1/e

PX/PY = sin²@
=1-cos²@
=1-1/e²
=e²-1/e²

 

[marsenal]

Originally posted by Fosweb
I have so far:

^PX² / _PY² = sin²@ [ ^{(a² + b²cos²@)(sin²@)} / _{(a²sin²@ + b²)(cos²@)} ]

Does the part in the [ ] brackets equal sin²@
Is this how you got the first bit marsenal?

Yeh thats preety much how I got it, with a lot of messing around with it, and boy am I not a big fan of algebra.

Originally posted by Constip8edSkunk
second part:
ae = asec@ -> sec@=e

Thanks! I guess I should have realised that myself. I got a focal chord mixed up with the latus rectum, so thats why I didn't make the connection.

 

[marsenal]

Originally posted by freaking_out
only rectangular hyperbolas.....it even says in ze sillabus.

Thanks