Home
Class 12
MATHS
The length of the Sub tangent at (2,2) t...

The length of the Sub tangent at `(2,2)` to the curve `x^5 = 2y^4` is

A

`5//2`

B

`8//5`

C

`2//5`

D

`5//8`

Text Solution

AI Generated Solution

The correct Answer is:
To find the length of the subtangent at the point (2, 2) to the curve given by the equation \( x^5 = 2y^4 \), we will follow these steps: ### Step 1: Understand the formula for the length of the subtangent The length of the subtangent at a point \((x_1, y_1)\) on the curve is given by the formula: \[ \text{Length of subtangent} = |y_1 \cdot \frac{dx}{dy}| \] ### Step 2: Differentiate the curve equation We start with the curve equation: \[ x^5 = 2y^4 \] To find \(\frac{dx}{dy}\), we differentiate both sides with respect to \(y\). Differentiating the left side: \[ \frac{d}{dy}(x^5) = 5x^4 \cdot \frac{dx}{dy} \] Differentiating the right side: \[ \frac{d}{dy}(2y^4) = 8y^3 \] Setting these equal gives us: \[ 5x^4 \cdot \frac{dx}{dy} = 8y^3 \] ### Step 3: Solve for \(\frac{dx}{dy}\) Rearranging the equation to solve for \(\frac{dx}{dy}\): \[ \frac{dx}{dy} = \frac{8y^3}{5x^4} \] ### Step 4: Substitute the point (2, 2) Now we substitute \(x = 2\) and \(y = 2\) into the expression for \(\frac{dx}{dy}\): \[ \frac{dx}{dy} = \frac{8(2)^3}{5(2)^4} = \frac{8 \cdot 8}{5 \cdot 16} = \frac{64}{80} = \frac{8}{10} = \frac{4}{5} \] ### Step 5: Calculate the length of the subtangent Now we can substitute \(y_1 = 2\) and \(\frac{dx}{dy} = \frac{4}{5}\) into the formula for the length of the subtangent: \[ \text{Length of subtangent} = |2 \cdot \frac{4}{5}| = \frac{8}{5} \] ### Final Answer Thus, the length of the subtangent at the point (2, 2) is: \[ \boxed{\frac{8}{5}} \] ---
Promotional Banner

Topper's Solved these Questions

  • TANGENTS AND NORMALS

    OBJECTIVE RD SHARMA ENGLISH|Exercise Chapter Test|25 Videos
  • TANGENTS AND NORMALS

    OBJECTIVE RD SHARMA ENGLISH|Exercise Section I - Solved Mcqs|49 Videos
  • SOLUTIONS OF TRIANGLES

    OBJECTIVE RD SHARMA ENGLISH|Exercise Exercise|20 Videos
  • THREE DIMENSIONAL COORDINATE SYSTEM

    OBJECTIVE RD SHARMA ENGLISH|Exercise Exercise|20 Videos

Similar Questions

Explore conceptually related problems

If the length of the tangent from (2,5) to the circle x^(2) + y^(2) - 5x +4y + k = 0 is sqrt(37) then find k.

The equation of the tangent at (2,\ 3) on the curve y^2=a x^3+b is y=4x-5 . Find the values of a and b .

The equation of the tangent at (2,3) on the curve y^2=a x^3+b is y=4x-5. Find the values of a and b

If the length of the tangent from (h,k) to the circle x^(2)+y^2=16 is twice the length of the tangent from the same point to the circle x^(2)+y^(2)+2x+2y=0 , then

If the length of the tangent from (5,4) to the circle x^(2) + y^(2) + 2ky = 0 is 1 then find k.

The length of the tangent from (0, 0) to the circle 2(x^(2)+y^(2))+x-y+5=0 , is

Find the length of sub-tangent to the curve y=e^(x//a)

The length of the sub-tangent to the hyperbola x^(2)-4y^(2)=4 corresponding to the normal having slope unity is (1)/(sqrtk), then the value of k is

If the length of sub-normal is equal to the length of sub-tangent at any point (3,4) on the curve y=f(x) and the tangent at (3,4) to y=f(x) meets the coordinate axes at Aa n dB , then the maximum area of the triangle O A B , where O is origin, is 45/2 (b) 49/2 (c) 25/2 (d) 81/2

If the length of sub-normal is equal to the length of sub-tangent at any point (3,4) on the curve y=f(x) and the tangent at (3,4) to y=f(x) meets the coordinate axes at Aa n dB , then the maximum area of the triangle O A B , where O is origin, is 45/2 (b) 49/2 (c) 25/2 (d) 81/2

OBJECTIVE RD SHARMA ENGLISH-TANGENTS AND NORMALS-Exercise
  1. The equation of the tangent to the curve y =x^(4) from the point (2,0...

    Text Solution

    |

  2. The point on the curve sqrt(x) + sqrt(y) = sqrt(a), the normal at whic...

    Text Solution

    |

  3. The length of the Sub tangent at (2,2) to the curve x^5 = 2y^4 is

    Text Solution

    |

  4. The angle between the curves y=sin x and y = cos x, 0 lt x lt (pi)/(2)...

    Text Solution

    |

  5. The line, which is parallel to X-axis and crosses the curve y = sqrtx ...

    Text Solution

    |

  6. A normal is drawn to parabola y^2=4ax at any point other than the vert...

    Text Solution

    |

  7. If the line a x+b y+c=0 is a normal to the curve x y=1, then a >0,b >...

    Text Solution

    |

  8. Show that the line d/a+y/b=1 touches the curve y=b e^(-x/a) at the poi...

    Text Solution

    |

  9. Find the euation of normal to the curve x=a( cos theta + theta sin th...

    Text Solution

    |

  10. The point P of the curve y^(2)=2x^(3) such that the tangent at P is p...

    Text Solution

    |

  11. Find the equation of tangents to the curve y=cos(x+y),-2pilt=xlt=2pi t...

    Text Solution

    |

  12. The equation of the tangents at the origin to the curve y^2=x^2(1+x) a...

    Text Solution

    |

  13. The coordinates of the points on the curve x=a(theta + sintheta), y=a(...

    Text Solution

    |

  14. The chord joining the points where x= p and x= q on the curve y= ax^2 ...

    Text Solution

    |

  15. Find the locus of point on the curve y^2=4a(x+asin (x/a)) where tangen...

    Text Solution

    |

  16. At what points on the curve y=x^2-4x+5 is the tangent perpendicu...

    Text Solution

    |

  17. The points of contact of the tangents drawn from the origin to the cur...

    Text Solution

    |

  18. If the area of the triangle included between the axes and any tangent ...

    Text Solution

    |

  19. The tangents to the curve x=a(theta - sin theta), y=a(1+cos theta) at ...

    Text Solution

    |

  20. The slope of the tangent to the curve y=sin^(-1) (sin x) " at " x=(3pi...

    Text Solution

    |