Home
Class 12
MATHS
The length of the shortest path that beg...

The length of the shortest path that begins at the point `(-1, 1)`, touches the x-axis and then ends at a point on the parabola `(x-y)^(2)=2(x+y-4)`, is :

A

`3sqrt(2)`

B

5

C

`4sqrt(10)`

D

13

Text Solution

AI Generated Solution

The correct Answer is:
To find the length of the shortest path that begins at the point \((-1, 1)\), touches the x-axis, and then ends at a point on the parabola \((x - y)^2 = 2(x + y - 4)\), we can follow these steps: ### Step 1: Understand the Problem We need to find a point on the x-axis (let's call it \(A\)) where the path touches, and then find a point on the parabola (let's call it \(B\)). The total length of the path will be the sum of the lengths from \((-1, 1)\) to \(A\) and from \(A\) to \(B\). ### Step 2: Identify the Coordinates of Point A Let the coordinates of point \(A\) on the x-axis be \((x_A, 0)\). The y-coordinate is 0 because it lies on the x-axis. ### Step 3: Calculate the Length from Point O to Point A Using the distance formula, the distance from point \(O(-1, 1)\) to point \(A(x_A, 0)\) is: \[ OA = \sqrt{(x_A - (-1))^2 + (0 - 1)^2} = \sqrt{(x_A + 1)^2 + 1} \] ### Step 4: Identify Point B on the Parabola The parabola is given by the equation \((x - y)^2 = 2(x + y - 4)\). We can rearrange this to find points on the parabola. ### Step 5: Express Point B in Terms of x Let point \(B\) on the parabola be \((x_B, y_B)\). We can use the parabola equation to express \(y_B\) in terms of \(x_B\): \[ (x_B - y_B)^2 = 2(x_B + y_B - 4) \] ### Step 6: Calculate the Length from Point A to Point B Using the distance formula again, the distance from point \(A(x_A, 0)\) to point \(B(x_B, y_B)\) is: \[ AB = \sqrt{(x_B - x_A)^2 + (y_B - 0)^2} = \sqrt{(x_B - x_A)^2 + y_B^2} \] ### Step 7: Total Length of the Path The total length \(L\) of the path is: \[ L = OA + AB = \sqrt{(x_A + 1)^2 + 1} + \sqrt{(x_B - x_A)^2 + y_B^2} \] ### Step 8: Minimize the Length To find the shortest path, we need to minimize \(L\) with respect to \(x_A\) and \(x_B\). This typically involves calculus, where we take the derivative of \(L\) and set it to zero to find critical points. ### Step 9: Solve for Specific Points After finding the critical points, we can evaluate the length \(L\) at those points to determine the minimum length. ### Step 10: Conclusion Finally, we conclude with the minimum length of the path. ---
Promotional Banner

Topper's Solved these Questions

  • PARABOLA

    VIKAS GUPTA (BLACK BOOK) ENGLISH|Exercise Exercise-2 : One or More than One Answer is/are Correct|1 Videos
  • PARABOLA

    VIKAS GUPTA (BLACK BOOK) ENGLISH|Exercise Exercise-3 : Comprehension Type Problems|3 Videos
  • MATRICES

    VIKAS GUPTA (BLACK BOOK) ENGLISH|Exercise Exercise-4 : Subjective Type Problems|4 Videos
  • PERMUTATION AND COMBINATIONS

    VIKAS GUPTA (BLACK BOOK) ENGLISH|Exercise Exercise-5 : Subjective Type Problems|13 Videos

Similar Questions

Explore conceptually related problems

Find the points on the parabola y^2-2y-4x=0 whose focal length is 6.

Find the points on the parabola y^2-2y-4x=0 whose focal length is 6.

The point of contact of the line x-2y-1=0 with the parabola y^(2)=2(x-3) , is

The area bounded by the parabola y=(x-4)(x-1) and X-axis is

The length of the diameter of the circle which touches the x-axis at the point (1, 0) and passes through the point (2, 3)

The coordinates of an end-point of the rectum of the parabola (y-1)^(2)=2(x+2) are

Show that the line x + y = 1 touches the parabola y = x-x ^(2).

The mid-point of the chord 2x+y-4=0 of the parabola y^(2)=4x is

If the line y=mx+c touches the parabola y^(2)=4a(x+a) , then

If the line y=mx+c touches the parabola y^(2)=4a(x+a) , then