Chord PQ subtending Right angle at centre of ellipse.

1. Apr 2, 2013

amk_dbz

1. The problem statement, all variables and given/known data

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)]

2. Relevant equations
Parametric form of points P(acos(θ),bsin(θ)),

3. 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.

Last edited by a moderator: Apr 4, 2013
2. Apr 3, 2013

haruspex

Theta is only a parameter here. It does not represent the angle the vector makes to the x axis.

3. Apr 3, 2013

amk_dbz

Oh! Interesting...I had failed to notice that and took it as the angle between positive x-axis and the position vector of the point.
Thanks

Ok, so how do I prove the property then ?

4. Apr 3, 2013

haruspex

let the points be parameterised by independent angles, then use the perpendicularity of the vectors to find the relationship between the two.

5. Apr 3, 2013

amk_dbz

Thanks again..
I tried solving it assuming parameters A and B [P(acosA,bsinA)] and [Q(acosB,bsinB)]
Therefore OP.OQ = a^2(cosA)(cosB) + b^2(sinA)(sinB) = 0

Now I simplified it in two ways leading to two equations
1) (b^2+a^2)[cos(A-B)]=(b^2-a^2)[cos(A+B)]

2) -(a^2)/(b^2) = tanAtanB

What should I do next?

Thanks.

6. Apr 4, 2013

haruspex

The LHS of the target equality is 1/[(OP)^2] + 1/[(OQ)^2], so obtain an expression for that in terms of the angles.

7. Apr 4, 2013

amk_dbz

I have been trying to simplify the equation but in vain....It is too messy and long. Please let me know if you have a better way.
Anyways thanks for assistance. :)

8. Apr 4, 2013

haruspex

Try substituting t = tan A, u = tan B, and converting all the trig functions into references to t and u. You can use tu = -a2/b2 to eliminate one of them.