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Hint: The intersection point satisfies both the equations of conic sections.

We are given the equations of two parabolas

${x^2} = 4y{\text{ (1)}}$

${y^2} = 4x{\text{ (2)}}$

We need to find the intersection point of the two parabolas other than the origin.

If the two parabolas intersect, above two equations should have common solutions.

Using equation (1) in equation (2), we get,

$

{\left( {\dfrac{{{x^2}}}{4}} \right)^2} = 4x \\

\Rightarrow {x^4} - 64x = 0 \\

\Rightarrow x\left( {{x^3} - {4^3}} \right) = 0 \\

$

Using identity ${a^3} - {b^3} = \left( {a - b} \right)\left( {{a^2} + {b^2} + ab} \right)$ in the above equation

$

\Rightarrow x\left( {x - 4} \right)\left( {{x^2} + 16 + 4x} \right) = 0 \\

\Rightarrow x = 0,x = 4,\left( {{x^2} + 16 + 4x} \right) = 0 \\

$

We neglect $\left( {{x^2} + 16 + 4x} \right) = 0$ as the roots of this equation are imaginary

Using $x = 4$ in equation (2), we get $y = 4, - 4$

Similarly, by using x=0 in equation (2), we get $y = 0$ which is the point of intersection already stated in the problem.

Now we need to check for the two obtained points, that is, $x = 4,y = 4$ and $x = 4,y = - 4$ for satisfaction of equation of (1)

Using $x = 4,y = - 4$ in equation (1), we get

$

{4^2} = 4 \times \left( { - 4} \right) \\

\Rightarrow LHS \ne RHS \\

$

Therefore point $x = 4,y = - 4$ is neglected as it does not satisfy equation (1)

Using $x = 4,y = 4$ in equation (1), we get

$

{4^2} = 4 \times \left( 4 \right) \\

\Rightarrow LHS = RHS \\

$

Since point $x = 4,y = 4$ satisfies both equation (1) and (2), $\left( {4,4} \right)$ is the other point of intersection of the above given parabolas.

Hence option C. $\left( {4,4} \right)$ is correct.

Note: Only real solutions are needed to be considered for the intersection of two conical sections. Also, it is advised to draw the figures in order to get an idea of points of intersection in order to solve the question in less time.

We are given the equations of two parabolas

${x^2} = 4y{\text{ (1)}}$

${y^2} = 4x{\text{ (2)}}$

We need to find the intersection point of the two parabolas other than the origin.

If the two parabolas intersect, above two equations should have common solutions.

Using equation (1) in equation (2), we get,

$

{\left( {\dfrac{{{x^2}}}{4}} \right)^2} = 4x \\

\Rightarrow {x^4} - 64x = 0 \\

\Rightarrow x\left( {{x^3} - {4^3}} \right) = 0 \\

$

Using identity ${a^3} - {b^3} = \left( {a - b} \right)\left( {{a^2} + {b^2} + ab} \right)$ in the above equation

$

\Rightarrow x\left( {x - 4} \right)\left( {{x^2} + 16 + 4x} \right) = 0 \\

\Rightarrow x = 0,x = 4,\left( {{x^2} + 16 + 4x} \right) = 0 \\

$

We neglect $\left( {{x^2} + 16 + 4x} \right) = 0$ as the roots of this equation are imaginary

Using $x = 4$ in equation (2), we get $y = 4, - 4$

Similarly, by using x=0 in equation (2), we get $y = 0$ which is the point of intersection already stated in the problem.

Now we need to check for the two obtained points, that is, $x = 4,y = 4$ and $x = 4,y = - 4$ for satisfaction of equation of (1)

Using $x = 4,y = - 4$ in equation (1), we get

$

{4^2} = 4 \times \left( { - 4} \right) \\

\Rightarrow LHS \ne RHS \\

$

Therefore point $x = 4,y = - 4$ is neglected as it does not satisfy equation (1)

Using $x = 4,y = 4$ in equation (1), we get

$

{4^2} = 4 \times \left( 4 \right) \\

\Rightarrow LHS = RHS \\

$

Since point $x = 4,y = 4$ satisfies both equation (1) and (2), $\left( {4,4} \right)$ is the other point of intersection of the above given parabolas.

Hence option C. $\left( {4,4} \right)$ is correct.

Note: Only real solutions are needed to be considered for the intersection of two conical sections. Also, it is advised to draw the figures in order to get an idea of points of intersection in order to solve the question in less time.

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