Prove that the curves x = y² and xy = k cut at right angles if 8k² = 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 = y² and xy = k

Putting x = y² in xy = k

y³ = 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 = y² with respect to x, we have:

1 = 2y dy/dx

⇒ dy/dx = 1/2y

Therefore, the slope of the tangent to the curve x = y² = 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

⇒ 8k² = 1

Hence, the given curves are cut at the right angle if 8k² = 1

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

Prove that the curves x = y² and xy = k cut at right angles if 8k² = 1

Summary:

Hence we have proved that the curves x = y² and xy = k cut at right angles if 8k² = 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