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# At what point of the curve

Question:

At what point of the curve $y=x^{2}$ does the tangent make an angle of $45^{\circ}$ with the $x$-axis?

Solution:

Given:

The curve is $y=x^{2}$

Differentiating the above w.r.t $\mathrm{x}$

$\Rightarrow y=x^{2}$

$\Rightarrow \frac{d y}{d x}=2 x^{2}-1$

$\Rightarrow \frac{d y}{d x}=2 x \ldots(1)$

$\therefore \frac{\mathrm{dy}}{\mathrm{dx}}=$ The Slope of the tangent $=\tan \theta$

Since, the tangent make an angle of $45^{\circ}$ with $x$ - axis

i.e,

$\Rightarrow \frac{d y}{d x}=\tan \left(45^{\circ}\right)=1 \ldots(2)$

$\therefore \tan \left(45^{\circ}\right)=1$

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

$\Rightarrow 2 x=1$

$\Rightarrow x=\frac{1}{2}$

Substituting $x=\frac{1}{2}$ in $y=x^{2}$, we get,

$\Rightarrow y=\left(\frac{1}{2}\right)^{2}$

$\Rightarrow y=\frac{1}{4}$

Thus, the required point is $\left(\frac{1}{2}, \frac{1}{4}\right)$