# Let C be a curve passing through M(2,2) such that the slope of the tangent at any point to the curve is reciprocal of the ordinate of the point. If the area bounded by curve C and line x=2 is A, then the value of `(3A)/(2)` is__.

Updated On: 27-4-2021

400+

000+

Answer

Text Solution

8

<img src="https://d10lpgp6xz60nq.cloudfront.net/physics_images/CEN_CAL_C09_E08_003_S01.png" width="80%"> <br> Let P(x,y) be any point on the curve C. <br>

"Now, "(dy)/(dx)=(1)/(y)

<br>

"or "ydy=dxrArr(y^(2))/(2)=x+k

<br> Since the curve passes through M(2,2), so k=0 <br>

rArr" "y^(2)=2x

<br>

therefore" Required area "=2overset(2)underset(0)intsqrt(2x)dx=2sqrt(2)xx(2)/(3)(x^(3//2))_(0)^(2)

<br>

=(4)/(3)sqrt(2)xx2sqrt(2)=(16)/(3)

sq. unit

Transcript

Time | Transcript |
---|---|

00:00 - 00:59 | hello everyone about today's question is let C be a passing through I am which continent is to such that the slope of the tangent of any point of the curve is a reciprocal of co-ordinate of a point if the area bounded by a Kursi and LINUX pulse X equals to 2 then the value of 3 upon to us so we have to find the value of 3.2 so late is the point and the coordinator of peace x y b a point of the Kursi so that divide upon X is equals to 1 by vi n separate the variables we get wider Y is equals to DX then apply integration integration |

01:00 - 01:59 | wicked y square upon 2 = 2 X + k k as any cost curve passing through two former to so we can put the value of x y so we get to square upon 2 equals to 2 + ke so the value of k = 2010 equation 1 so why is Y squared equals to X and the required area equals to 2022 root 2 x dx after |

02:00 - 02:59 | solving this we get 2 root 2 into 2 by 3 x to the power 3 by 2 limit 0 to 2 so after putting the limits we get 4 root 2 upon 3 into 2 root 2 and the value of a 16 upon 3 square unit and the value of 3.2 is 16 into two 3.3 into two so it's 8 square unit it is our final answer thank you |

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