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amk_dbz
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Homework Statement
If P and Q are two points on ellipse [(x^2/a^2)+(y^2/b^2)]=1 such that PQ subtends a right angle at the centre O then.Prove that 1/[(OP)^2] + 1/[(OQ)^2] = [1/(a^2)] +[1/(b^2)]
Homework Equations
Parametric form of points P(acos(θ),bsin(θ)),
The Attempt at a Solution
Since the angle between them is 90 degrees therefore the points are P[acos(θ),bsin(θ)]
and Q[acos(90+θ),bsin(90+θ)] = Q[-asin(θ),bcos(θ)]
FirstNow I tried the distance formula but in vain
SecondI also tried using vectors as follows
Position vector of P= acos(θ) i +bsin(θ) j
Q= -asin(θ)i +bcos(θ) j
Since OP and OQ are perpandicular their dot product is zero
OP.OQ = 0
acos(θ).[-asin(θ)] + bsin(θ).[bcoz(θ)]=0
b^2[sin(θ)cos(θ)]- a^2[sin(θ)cos(θ) = 0
which gives a^2=b^2
a = b
I have 2 questions, 1)How to prove this?
2)Why does the vector method give a=b (which i think is only a special case). Did I do anything wrong?
Any help will be appreciated...Thanks.
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