How Do You Prove Point Equidistance and Find Loci in Geometry?

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To prove that a point P(x,y) is equidistant from points A(a+b,b-a) and B(a-b,a+b), it is established that bx = ay. The discussion also addresses finding the locus of a variable point Z(a cos(θ), b sin(θ)), where θ is variable, leading to the conclusion that the locus can be represented by the equation x²/a² + y²/b² = 1. Participants noted the challenges of using the distance formula for these proofs and suggested alternative approaches. Ultimately, the solution for the locus was derived without the distance formula, confirming the geometric relationships involved.
Kartik.
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1. A point P(x,y) is given equidistant from the points A(a+b,b-a) and B(a-b,a+b), then prove that bx = ay
also find the locus of the variable point Z(a cos (theta), b sin (theta)), where (theta) is a variable quantity.




2. T0 prove that ax = by



3. In an attempt towards implementing what's given, i applied the distance formula knowing the fact that it will be a lengthy solution but still i did not came up with an answer.
and the second part of the question, i am still doubtful about how to find a locus to a variable point. The value of theta is variable so, the x-coordinate and y-coordinate will oscillate from [-1,1].
 
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For the first one, could you show us what you've done? You can skip some trivial working out steps to save yourself time if you like, but the important parts would be what equality you set up by using the distance formula, and what answer you arrived at at the end.

For the second, it's quite similar to solving the locus of a circle (x,y)=(a\cos\theta,a\sin\theta) except now in this case we have a b constant as well, so simply using x^2+y^2=a^2 won't work.

But what about \frac{x^2}{a^2} can you see a way of using this and doing something similar for the y variable?
 
Solved 'em all. Thank you.
Not with the distance formula but deriving the locus of the point equidistant from A and B. for the second one its x^2/a^2+y^2/b^2 = 1
 
Since ##px^9+q## is the factor, then ##x^9=\frac{-q}{p}## will be one of the roots. Let ##f(x)=27x^{18}+bx^9+70##, then: $$27\left(\frac{-q}{p}\right)^2+b\left(\frac{-q}{p}\right)+70=0$$ $$b=27 \frac{q}{p}+70 \frac{p}{q}$$ $$b=\frac{27q^2+70p^2}{pq}$$ From this expression, it looks like there is no greatest value of ##b## because increasing the value of ##p## and ##q## will also increase the value of ##b##. How to find the greatest value of ##b##? Thanks
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