Conics question (1 Viewer)

[Wohzazz]

P(asec@, btan@) lies on the hyperbola x*2/a*2-y*2/b*2=1. The tangent at P cuts the x-axis at X and y-axis at Y. Show that PX/PY=sin*2@.

IS there any short cuts instead of manipulating super long equations that you have so much chance of getting wrong?

 

[spice girl]

i believe that the equation of the tangent is the only one u need, and the one that u shouldn't really get wrong.

And as for PX/PY, u can use similar triangles to make another shortcut. drop a perpendicular from P onto the X axis and call that Q. then becos OXY is similar to QXP, u can say PX/PY = QX/QO

u know X is the pt (acos@, 0) from solving the tangent eqn at y=0, and Q is the pt (asec@, 0)

jus need to prove that (asec@-acos@)/asec@ = sin^2@

 

[Wohzazz]

Wow geometry. Nicely done. I should try to do other conics question geometricly. Some of them, the equations becomes so ridiculous.

Also, did you know the values, say for X and Y from heart? You don't need to remember it right because you always need to derive it?