# Find the points on the curve

Question:

Find the points on the curve $x y+4=0$ at which the tangents are inclined at an angle of $45^{\circ}$ with the $x$-axis.

Solution:

Let the required point be (x1y1).
Slope of the tangent at this point = tan 45°">°° = 1
Given:

$x y+4=0 \ldots(1)$

Since the point satisfies the above equation,

$x_{1} y_{1}+4=0 \quad \ldots(2)$

On differentiating equation (2) both sides with respect to $x$, we get

$x \frac{d y}{d x}+y=0$

$\Rightarrow \frac{d y}{d x}=\frac{-y}{x}$

Slope of the tangent at $\left(x_{1}, y_{1}\right)=\left(\frac{d y}{d x}\right)_{(x, y)}=\frac{-y_{1}}{x_{1}}$

Slope of the tangent $=1$ [Given]

$\therefore \frac{-y_{1}}{x_{1}}=1$

$\Rightarrow x_{1}=-y_{1}$

On substituting the value of $x_{1}$ in eq. $(2)$, we get

$-y_{1}^{2}+4=0$

$\Rightarrow y_{1}^{2}=4$

$\Rightarrow y_{1}=\pm 2$

Case 1

When $y_{1}=2, x_{1}=-y_{1}=-2$

$\therefore\left(x_{1}, y_{1}\right)=(-2,2)$

Case 2

When $y_{1}=-2, x_{1}=-y_{1}=2$

$\therefore\left(x_{1}, y_{1}\right)=(2,-2)$