Conics (1 Viewer)

GaganDeep

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Hey, i got this question, COuld you also please post a diagram aswell. THanks inadvandce

7) P(acos(theta),bsin (theta) and Q (acos (thai),b sin (thai) lie on the ellipse (normal one)
if PQ subtends a right angle at (0,0) show that tan (theta)tan(thai)= -b^2/a^2
b) ud PQ subtends a right angle at (a,0) show that tan(theta/2)tan(thai/2)= -b^2/a^2
Thanks
 

haboozin

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GaganDeep said:
Hey, i got this question, COuld you also please post a diagram aswell. THanks inadvandce

7) P(acos(theta),bsin (theta) and Q (acos (thai),b sin (thai) lie on the ellipse (normal one)
if PQ subtends a right angle at (0,0) show that tan (theta)tan(thai)= -b^2/a^2
b) ud PQ subtends a right angle at (a,0) show that tan(theta/2)tan(thai/2)= -b^2/a^2
Thanks
the first part is very easy

what do you think about when it says right angle

m1.m2 = -1

so basicly b^2/a^2cot@cot& = -1
so -b^2/a^2 = tan@tan&

b. this ones a bit more difficault, since its not exactly 0,0
but if u get the gradient for P and Q and do the same thing as above you get the answer..




i used to never believe in diagrams, but it helped me with conics, so maybe you sohuld draw diagrams
 

c0okies

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haboozin.. for question 1.. isnt that formula for the gradient the gradient of a tangent?
 

Riviet

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Remember that if m1m2=-1, then it means that the two gradients, m1 and m2 are perpendicular, hence the right angle.
 

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