Home
Class 12
MATHS
y^2=4ax is a parabola . A line segment j...

`y^2=4ax` is a parabola . A line segment joining focus of parabola to any moving point `(at^2,2at)` is made . Then locus of mid-point of the line segment is a parabola with the directrix

A

`x=a/2`

B

`x=0`

C

`x=-a/2`

D

`x=a`

Text Solution

AI Generated Solution

The correct Answer is:
To solve the problem, we need to find the locus of the midpoint of the line segment joining the focus of the parabola \( y^2 = 4ax \) to a moving point on the parabola given by \( P(at^2, 2at) \). ### Step-by-Step Solution: 1. **Identify the Focus of the Parabola**: The standard form of the parabola \( y^2 = 4ax \) has its focus at the point \( (a, 0) \). 2. **Define the Moving Point**: The moving point \( P \) on the parabola can be represented as \( P(at^2, 2at) \). 3. **Find the Midpoint**: Let the midpoint \( Q \) of the line segment joining the focus \( S(a, 0) \) and the point \( P(at^2, 2at) \) be given by: \[ Q\left( \frac{a + at^2}{2}, \frac{0 + 2at}{2} \right) = Q\left( \frac{a(1 + t^2)}{2}, at \right) \] 4. **Express Coordinates of Midpoint**: Let the coordinates of the midpoint \( Q \) be \( (h, k) \): \[ h = \frac{a(1 + t^2)}{2}, \quad k = at \] 5. **Eliminate the Parameter \( t \)**: From the equation \( k = at \), we can express \( t \) as: \[ t = \frac{k}{a} \] Substituting this into the equation for \( h \): \[ h = \frac{a\left(1 + \left(\frac{k}{a}\right)^2\right)}{2} = \frac{a + \frac{k^2}{a}}{2} = \frac{a^2 + k^2}{2a} \] 6. **Rearranging the Equation**: Rearranging the equation gives: \[ 2ah = a^2 + k^2 \] This can be rewritten as: \[ k^2 = 2ah - a^2 \] 7. **Identify the Locus**: The equation \( k^2 = 2ah - a^2 \) represents a parabola with its vertex at \( (a, 0) \) and opens to the right. 8. **Find the Directrix**: The directrix of a parabola in the form \( k^2 = 4px \) is given by \( x = -p \). Here, we can see that \( 2a \) corresponds to \( 4p \), thus \( p = \frac{a}{2} \). Therefore, the directrix is: \[ x = -\frac{a}{2} \] ### Final Answer: The locus of the midpoint of the line segment is a parabola with the directrix given by: \[ x = -\frac{a}{2} \]

To solve the problem, we need to find the locus of the midpoint of the line segment joining the focus of the parabola \( y^2 = 4ax \) to a moving point on the parabola given by \( P(at^2, 2at) \). ### Step-by-Step Solution: 1. **Identify the Focus of the Parabola**: The standard form of the parabola \( y^2 = 4ax \) has its focus at the point \( (a, 0) \). 2. **Define the Moving Point**: ...
Promotional Banner

Topper's Solved these Questions

  • JEE MAIN 2021

    JEE MAINS PREVIOUS YEAR|Exercise SECTION-A|100 Videos
  • JEE MAIN 2021

    JEE MAINS PREVIOUS YEAR|Exercise SECTION-B|50 Videos
  • JEE MAIN

    JEE MAINS PREVIOUS YEAR|Exercise QUESTION|1 Videos
  • JEE MAIN 2022

    JEE MAINS PREVIOUS YEAR|Exercise Question|454 Videos

Similar Questions

Explore conceptually related problems

If P is point on the parabola y ^(2) =8x and A is the point (1,0), then the locus of the mid point of the line segment AP is

A tangent to the parabola y^(2)=4ax meets axes at A and B .Then the locus of mid-point of line AB is

If any tangent to the parabola x^(2)=4y intersects the hyperbola xy=2 at two points P and Q , then the mid point of the line segment PQ lies on a parabola with axs along

The locus of the mid-point of the line segment joning the focus to a moving point on the parabola y^(2)=4ax is another parabola with directrix

The mid-point of the line segment joining the points (- 2, 4) and (6, 10) is

The locus of the middle points of the focal chord of the parabola y^(2)=4ax , is

If any tangent to the parabola x^(2) = 4y intersects the hyperbola xy = 2 at two points P and Q, then the mid point of line segment PQ lies on a parabola with axis along: