Question

Find a point on the curve xy + 4 = 0 at which the tangents are inclined at an angle of 45° with the x–axis.

Answer 1

Given:

If a tangent line to the curve y = f(x) makes an angle θ with x – axis in the positive direction, then \(\frac{dy}{dx}\)=The Slope of the tangent = tanθ

xy + 4 = 0

Differentiating the above w.r.t x

⇒ x x \(\frac{d}{dx}\) (y) + y x \(\frac{d}{dx}\) (x) + \(\frac{d}{dx}\) (4) = 0

⇒ x \(\frac{dy}{dx}\)+ y = 0

⇒ x \(\frac{dy}{dx}\)= – y

⇒ \(\frac{dy}{dx}\) = \(\frac{-y}{x}\)...(1)

Also, \(\frac{dy}{dx}\) = tan45° = 1 ...(2)

From (1) & (2),we get,

⇒ \(\frac{-y}{x}\) = 1

⇒ x = – y

Substitute in xy + 4 = 0,we get

⇒ x( – x) + 4 = 0

⇒ – x² + 4 = 0

⇒ x² = 4

⇒ x = \(\pm\)2

so when x = 2,y = – 2

& when x = – 2,y = 2

Thus, the points are (2, – 2) & ( – 2,2)