Question

The value of k^2 if y^2=4x and xy=k cut orthogonally is

Answer 1

Hi Mate!!



Two curves cuts orthogonally if the product of their slopes =-1

Slope of ist curve m = 2 / y


Slope of second curve m' = -y / x

=>. m × m' = { 2 / y } × { - y / x } = -1

=>. 2 / x = 1

=>. 2 ×4 / k² = 1

=>. k² = 8

Answer 2

The value of k² is 32.

Given:

y² = 4x and x*y =k cut orthogonally.

To Find:

The value of K²

Solution:

y²= 4x

Differentiating both sides  with respect to x

2* y*dy/dx = 4

dy/dx = 2/y

⇒ The slope of y²= 4x (m1) = 2/y

x*y =k

Differentiating both sides  with respect to x

x*dy/dx + y = 0

dy/dx= -y/x

⇒ The slope of x*y=k (m2) = -y/x

Product of slopes when they cut orthogonally is -1

⇒ m1 *m2 = -1

⇒ 2/y * -y/x = -1

⇒-2/x=-1

⇒ x= 2

Substituting x= 2 in equation x*y =k

2y=k

squaring on both sides

4y²=k² -------------------------(1)

In equation y²=4x put x=2

y²=4*2

y² = 8

Substituting y² value in equation (1)

4*8=k²

⇒ k² = 32

∴ The value of k² is 32.

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