The two parabolas ${x^2} = 4y$ and ${y^2} = 4x$ meet in two distinct points. One of these is the origin and the other is:
A. $\left( {2,2} \right)$
B. $\left( {4, - 4} \right)$
C. $\left( {4,4} \right)$
D. $\left( { - 2,2} \right)$
Last updated date: 20th Mar 2023
•
Total views: 309.3k
•
Views today: 7.87k
Answer
309.3k+ views
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.
Recently Updated Pages
If ab and c are unit vectors then left ab2 right+bc2+ca2 class 12 maths JEE_Main

A rod AB of length 4 units moves horizontally when class 11 maths JEE_Main

Evaluate the value of intlimits0pi cos 3xdx A 0 B 1 class 12 maths JEE_Main

Which of the following is correct 1 nleft S cup T right class 10 maths JEE_Main

What is the area of the triangle with vertices Aleft class 11 maths JEE_Main

KCN reacts readily to give a cyanide with A Ethyl alcohol class 12 chemistry JEE_Main

Trending doubts
What was the capital of Kanishka A Mathura B Purushapura class 7 social studies CBSE

Difference Between Plant Cell and Animal Cell

Write an application to the principal requesting five class 10 english CBSE

Ray optics is valid when characteristic dimensions class 12 physics CBSE

Give 10 examples for herbs , shrubs , climbers , creepers

Tropic of Cancer passes through how many states? Name them.

Write the 6 fundamental rights of India and explain in detail

Write a letter to the principal requesting him to grant class 10 english CBSE

Name the Largest and the Smallest Cell in the Human Body ?
