Home
Class 12
MATHS
Find a point on the graph of y=x^(3), wh...

Find a point on the graph of `y=x^(3)`, where the tangent is parallel to the chord joining (1, 1) and (3, 27).

Text Solution

AI Generated Solution

The correct Answer is:
To find a point on the graph of \( y = x^3 \) where the tangent is parallel to the chord joining the points \( (1, 1) \) and \( (3, 27) \), we can follow these steps: ### Step 1: Find the slope of the chord The slope of the chord joining the points \( (1, 1) \) and \( (3, 27) \) can be calculated using the formula: \[ \text{slope} = \frac{y_2 - y_1}{x_2 - x_1} \] Here, \( (x_1, y_1) = (1, 1) \) and \( (x_2, y_2) = (3, 27) \). Calculating the slope: \[ \text{slope} = \frac{27 - 1}{3 - 1} = \frac{26}{2} = 13 \] ### Step 2: Find the derivative of the function The derivative of the function \( y = x^3 \) gives us the slope of the tangent at any point on the curve. \[ \frac{dy}{dx} = 3x^2 \] ### Step 3: Set the derivative equal to the slope of the chord To find the points where the tangent is parallel to the chord, we set the derivative equal to the slope of the chord: \[ 3x^2 = 13 \] ### Step 4: Solve for \( x \) Now, we solve for \( x \): \[ x^2 = \frac{13}{3} \] Taking the square root of both sides: \[ x = \pm \sqrt{\frac{13}{3}} \] ### Step 5: Find the corresponding \( y \) values Now we need to find the corresponding \( y \) values for both \( x \) values using the original function \( y = x^3 \). 1. For \( x = \sqrt{\frac{13}{3}} \): \[ y = \left(\sqrt{\frac{13}{3}}\right)^3 = \frac{13\sqrt{\frac{13}{3}}}{3} \] 2. For \( x = -\sqrt{\frac{13}{3}} \): \[ y = \left(-\sqrt{\frac{13}{3}}\right)^3 = -\frac{13\sqrt{\frac{13}{3}}}{3} \] ### Step 6: Write the points Thus, the points on the graph where the tangent is parallel to the chord are: \[ \left(\sqrt{\frac{13}{3}}, \frac{13\sqrt{\frac{13}{3}}}{3}\right) \quad \text{and} \quad \left(-\sqrt{\frac{13}{3}}, -\frac{13\sqrt{\frac{13}{3}}}{3}\right) \] ### Summary of the solution The points on the graph of \( y = x^3 \) where the tangent is parallel to the chord joining \( (1, 1) \) and \( (3, 27) \) are: 1. \( \left(\sqrt{\frac{13}{3}}, \frac{13\sqrt{\frac{13}{3}}}{3}\right) \) 2. \( \left(-\sqrt{\frac{13}{3}}, -\frac{13\sqrt{\frac{13}{3}}}{3}\right) \)
Promotional Banner

Topper's Solved these Questions

  • APPLICATIONS OF DERIVATIVES

    MODERN PUBLICATION|Exercise EXERCISE 6 (d) (Long Answer Type Questions (I))|46 Videos
  • APPLICATIONS OF DERIVATIVES

    MODERN PUBLICATION|Exercise EXERCISE 1 (e) (Short Answer Type Questions)|16 Videos
  • APPLICATIONS OF DERIVATIVES

    MODERN PUBLICATION|Exercise EXERCISE 6 (c) (Long Answer Type Questions (I))|42 Videos
  • APPLICATIONS OF THE INTEGRALS

    MODERN PUBLICATION|Exercise CHAPTER TEST|12 Videos

Similar Questions

Explore conceptually related problems

Find a point on the curve y=x^3+1 where the tangent is parallel to the chord joining (1,\ 2) and (3,\ 28) .

Find a point on the curve y=x^(3)-3x where the tangent is parallel to the chord joining (1,-2) and (2,2) .

Find a point on the parabola y=(x-3)^2 , where the tangent is parallel to the chord joining (3, 0) and (4, 1).

Find a point on the curve y=x^(2)+x, where the tangent is parallel to the chord joining (0,0) and (1,2).

Find a point on the curve y=x^3-3x where the tangent is parallel to the chord joining (1,\ -2) and (2,\ 2) .

Find a point on the curve y=x^2+x , where the tangent is parallel to the chord joining (0, 0) and (1, 2).

Find a point on the parabola y=(x-4)^2 , where the tangent is parallel to the chord joining (4, 0) and (5, 1).

Find a point on the parabola y=(x-2)^(2) , where the tangent is parallel to the chord joining (2, 0) and (4, 4).

Find a point on the parabola y = (x -3)^(2) , where the tangent is parallel to the chord joining (3, 0) and (4, 1)

Find a point on the curve y=x^(3)-3x where tangent is parallel to the chord joining (1,-2) and (2,7)