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...

#Algebra #Geometry #Parabola

Easy Math Editor

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Comments

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

Aritra Jana - 6 years, 7 months ago

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

Vishal Ramesh - 6 years, 7 months ago

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)

Indraneel Mukhopadhyaya - 5 years, 9 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$