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
3 years, 7 months ago

MarkdownAppears as
*italics* or _italics_ italics
**bold** or __bold__ bold
- bulleted- list
• bulleted
• list
1. numbered2. list
1. numbered
2. list
Note: you must add a full line of space before and after lists for them to show up correctly
paragraph 1paragraph 2

paragraph 1

paragraph 2

[example link](https://brilliant.org)example link
> This is a quote
This is a quote
    # I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
# I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
MathAppears 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} $$\boxed{123}$$

Sort by:

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)

- 2 years, 9 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...

- 3 years, 7 months ago

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

- 3 years, 7 months ago