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The locus of the point (a cos³ q, a sin³ q) is

x^2/3 – y^2/3 = a^2/3
x^2/3 + y^2/3 = a^2/3
x^2/3 + y^2/3 = a^3/2
x^3/2 + y^2/3 = a^3/2

hint Hint:

Here we have to give the locus of the (a cos³ q, a sin³ q). Here x and y point are given so put this value in options and find which one satisfy the equation.

The correct answer is: x^2/3 + y^2/3 = a^2/3

Here we have to find the correct equation where locus of the point is given.
Firstly, we have locus which is (a cos³ q, a sin³ q)
So, x = a cos³q and y = a sin³q
We have, (1)
$x to the power of 3 over 2 end exponent$ + $y to the power of 2 over 3 end exponent$ = $a to the power of 3 over 2 end exponent$
LHS
=> (a cos³ q) $3 over 2$ + (a sin³ q) $3 over 2$
=> ( $a to the power of 3 over 2 end exponent$ $cos to the power of 9 over 2 end exponent$ q) + ( $a to the power of 3 over 2 end exponent$ sin² q)
=> $a to the power of 3 over 2 end exponent$ (( $cos to the power of 9 over 2 end exponent$ q) + (sin² q))
So, it is not equal to RHS, it is wrong option
(2)
$x to the power of 2 over 3 end exponent$ + $y to the power of 2 over 3 end exponent$ = $a to the power of 3 over 2 end exponent$
LHS =
=> $left parenthesis a space cos 3 q space right parenthesis to the power of 2 over 3 end exponent$ + $space left parenthesis a space sin 3 q space right parenthesis to the power of 2 over 3 end exponent$
=> ( $a to the power of 2 over 3 end exponent$ cos² q ) + ( $a to the power of 2 over 3 end exponent$ sin² q )
=> $a to the power of 2 over 3 end exponent$ (cos² q ) + (sin² q )
=> $a to the power of 2 over 3 end exponent$ x 1 [ since, (cos² x ) + (sin² x ) = 1 ]
=> $a to the power of 2 over 3 end exponent$
It is not equal to RHS, it is wrong option.
(3)
$x to the power of 2 over 3 end exponent$ + $y to the power of 2 over 3 end exponent$ = $a to the power of 2 over 3 end exponent$
LHS=
=> $left parenthesis a c o s 3 q right parenthesis to the power of 2 over 3 end exponent$ - $space left parenthesis a space sin 3 q space right parenthesis to the power of 2 over 3 end exponent$
=> ( $a to the power of 2 over 3 end exponent$ cos² q) - ( $a to the power of 2 over 3 end exponent$ sin² q)
=> $a to the power of 2 over 3 end exponent$ (cos² q) -(sin² q)
It is not equal to RHS, it is wrong option.
(4)
$x to the power of 2 over 3 end exponent$ + $y to the power of 2 over 3 end exponent$ = $a to the power of 2 over 3 end exponent$
LHS=
=> $left parenthesis a c o s cubed q right parenthesis to the power of 2 over 3 end exponent$ + $left parenthesis a sin cubed q right parenthesis to the power of 2 over 3 end exponent$
=> ( $a to the power of 2 over 3 end exponent$ cos² q) + ( $a to the power of 2 over 3 end exponent$ sin² q)
=> $a to the power of 2 over 3 end exponent$ (cos² q) + (sin² q)
=> $a to the power of 2 over 3 end exponent$ x 1 [ since, (cos² x) + (sin² x) = 1]
=> $a to the power of 2 over 3 end exponent$
Here, LHS = RHS, so it is the correct answer.
Therefore, the correct answer is $x to the power of 2 over 3 end exponent$ + $y to the power of 2 over 3 end exponent$ = $a to the power of 2 over 3 end exponent$

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The locus of the point (a cos3 q, a sin3 q) is

x2/3 – y2/3 = a2/3x2/3 + y2/3 = a2/3x2/3 + y2/3 = a3/2x3/2 + y2/3 = a3/2

Here we have to give the locus of the (a cos3 q, a sin3 q). Here x and y point are given so put this value in options and find which one satisfy the equation.

The correct answer is: x2/3 + y2/3 = a2/3

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The locus of the point (a cos³ q, a sin³ q) is

x^2/3 – y^2/3 = a^2/3
x^2/3 + y^2/3 = a^2/3
x^2/3 + y^2/3 = a^3/2
x^3/2 + y^2/3 = a^3/2

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The correct answer is: x^2/3 + y^2/3 = a^2/3

The are of the triangle inscribed in the parabola y² = 4x, the ordinates of whose vertices are 1, 2 and 4 is

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The line x + my + n = 0 will be a tangent to the parabola y² = 4ax if

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