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# Can someone tell me how to approach and solve the problem

Parabolas $y^2=4a(x-c)$ and $x^2=4a(y-c')$ where c and c' are variables and touch each other. Locus of their point of contact is :- $a)xy=a^2$ $b)xy=2a^2$ $c)xy=4a^2$ d) None of these

Please post the solution in DETAIL...

Note by Parag Zode
2 years, 9 months ago

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Slope of tangent at point of contact should be same for both the curves.Hence for both the curves take derivative of y with respect to x, substitute the point of contact in the expression obtained by the differentiation and equate.You will then get the required locus as xy=4* (square of a) · 1 year, 11 months ago

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You can find y in terms of x, a, andc' from eqn.2 . Now substitute it in the first eqn to get the soln... · 2 years, 9 months ago

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can you clarify something please:

if $$c$$ and $$c'$$ are variables, wouldn't the locus correspond to a 3 variable equation?

otherwise, considering them as constants, i get $$c$$ as answer · 2 years, 9 months ago

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