Conics (1 Viewer)

[mazza_728]

P(asec@, btan@) and Q(asec(-@), btan(-@)) are the extremities of thelatus rectum x=ae of the hyperbola x^2/a^2-y^2/b^2 = 1 Show that
a) sec@=e
b) PQ has a length 2 b^2/a

Dont understand it at all??
Can anyone explain it for me?
Thanks xoxo

 

[Supra]

P lies on x=ae coz it says its the extremity of hte latus rectum...
therefore asec@=ae therefore sec@=e

the length of PQ =2btan@

we know that b²=a²(e²-1) (property of hyperbola)
well from part 1 sec@=e
therefore b²=a²(sec²@-1)
b²=a²tan²@
tan²@=b²/a²
tan@=b/a (+/-)
therfore PQ=2b x b/a (PQ is a length therfore it has to b positive)

then u get required result

 

[paganinio]

for hyperbola x^2/a^2-y^2/b^2 = 1(a,b>0), the length of latus rectum is always
L=2*a(e^2-1)=2*a*e^2-2*a=(2*a*c^2)/(a^2)-(2*a^3)/(a^2)=2*a(c^2-a^2)/(a^2)=2*a*(b^2)/(a^2)=2*(b^2)/a