Find a point on the curve
Question:

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

Solution:

Given:

The curve is $x y+4=0$

If a tangent line to the curve $y=f(x)$ makes an angle $\boldsymbol{\theta}$ with $x$-axis in the positive direction, then

$\frac{\mathrm{d} y}{\mathrm{dx}}=$ The Slope of the tangent $=\tan \theta$

$x y+4=0$

Differentiating the above w.r.t $x$

$\Rightarrow x \times \frac{d}{d x}(y)+y \times \frac{d}{d x}(x)+\frac{d}{d x}(4)=0$

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

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

$\Rightarrow \frac{d y}{d x}=\frac{-y}{x} \ldots(1)$

Also, $\frac{\mathrm{dy}}{\mathrm{dx}}=\tan 45^{\circ}=1 \ldots(2)$

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

$\Rightarrow \frac{-y}{x}=1$

$\Rightarrow x=-y$

Substitute in $x y+4=0$, we get

$\Rightarrow x(-x)+4=0$

$\Rightarrow-x^{2}+4=0$

$\Rightarrow x^{2}=4$

$\Rightarrow x=\pm 2$

so when $x=2, y=-2$

$\&$ when $x=-2, y=2$

Thus, the points are $(2,-2) \&(-2,2)$

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