Question

Find the points on the curve xy + 4 = 0 at which the tangents are inclined at an angle of 45° with the x-axis ?

Answer 1

Let the required point be (x₁, y₁).
Slope of the tangent at this point = tan 45° 

Given :

\[xy + 4 = 0 . . . \left( 1 \right)\]

\[\text { Since the point satisfies the above equation}, \]

\[ x_1 y_1 + 4 = 0 . . . \left( 2 \right)\]

\[\text { On differentiating equation }\left( 2 \right)\text { both sides with respect tox, we get } \]

\[x\frac{dy}{dx} + y = 0\]

\[ \Rightarrow \frac{dy}{dx} = \frac{- y}{x}\]

\[\text { Slope of the tangent at }\left( x_1 , y_1 \right)= \left( \frac{dy}{dx} \right)_\left( x, y \right) = \frac{- y_1}{x_1}\]

\[\text { Slope of the tangent =1 [Given]}\]

\[ \therefore \frac{- y_1}{x_1} = 1\]

\[ \Rightarrow x_1 = - y_1 \]

\[\text { On substituting the value of } x_1 \text {in eq. (2), we get }\]

\[ - {y_1}^2 + 4 = 0\]

\[ \Rightarrow {y_1}^2 = 4\]

\[ \Rightarrow y_1 = \pm 2\]

\[\text { Case} 1\]

\[\text { When }y_1 = 2, x_1 = - y_1 = - 2\]

\[\therefore ( x_1 , y_1 ) = (-2, 2)\]

\[\text { Case } 2\]

\[\text { When }y_1 = - 2, x_1 = - y_1 = 2\]

\[\therefore\left( x_1 , y_1 \right)= (2, -2)\]