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# Prove that the curves x = y^{2 }and xy = k cut at right angles if 8k^{2} = 1

Hint: Two curves intersect at right angle if the tangents to the curve at the point of intersection are perpendicular to each other

**Solution:**

The equations of the given curves are

x = y^{2 }and xy = k

Putting x = y^{2 }in xy = k

y^{3} = k

⇒ y = k^{1/3}

⇒ x = k^{2/3}

Thus, the point of intersection of the given curves is (k^{2/3}, k^{1/3})

Differentiating x = y^{2 }with respect to x, we have:

1 = 2y dy/dx

⇒ dy/dx = 1/2y

Therefore, the slope of the tangent to the curve x = y^{2} = at (k^{2/3}, k^{1/3}) is

dy/dx](k^{2/3}, k^{1/3}) = 1/2k^{1/3}

On differentiating xy = k with respect to x, we have:

x dy/dx + y = 0

⇒ dy/dx = - y/x

Now, slope of the tangent to the curve xy = k at (k^{2/3}, k^{1/3}) is

dy/dx](k^{2/3}, k^{1/3}) = - y/x](k^{2/3}, k^{1/3})

= - k^{1/3}/k^{2/3}

= - 1/k^{1/3}

We know that two curves intersect at right angles if the tangents to the curves at the point of intersection i.e., at (k^{2/3}, k^{1/3}) are perpendicular to each other.

This implies that we should have the product of the tangents as - 1.

Thus, the given two curves are cut at right angles if the product of the slopes of their respective tangents at (k^{2/3}, k^{1/3}) is - 1.

⇒ (1/2k^{1/3})(- 1/k^{1/3}) = - 1

⇒ 2k^{1/3 }= 1

⇒ (2k^{2/3})^{3 }= (1)^{3}

⇒ 8k^{2} = 1

Hence, the given curves are cut at the right angle if 8k^{2} = 1

NCERT Solutions Class 12 Maths - Chapter 6 Exercise 6.3 Question 23

## Prove that the curves x = y^{2 }and xy = k cut at right angles if 8k^{2} = 1

**Summary:**

Hence we have proved that the curves x = y^{2 }and xy = k cut at right angles if 8k^{2} = 1.The slope of a line is nothing but the change in y coordinate with respect to the change in x coordinate of that line

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