Find the co-ordinates of the point on the curve

Solution:

Equation of the curve is given by, √x + √y = 4

Now, let (x1, y1) be he required point on the curve

So, √x1 + √y1 = 4

On differentiating on both the sides w.r.t. x1, we get

$\frac{d}{d x_{1}} \sqrt{x_{1}}+\frac{d}{d x_{1}} \sqrt{y_{1}}=\frac{d}{d x_{1}}(4)$

$\frac{1}{2 \sqrt{x_{1}}}+\frac{1}{2 \sqrt{y_{1}}} \cdot \frac{d y_{1}}{d x_{1}}=0$

$\Rightarrow \frac{1}{\sqrt{x_{1}}}+\frac{1}{\sqrt{y_{1}}} \cdot \frac{d y_{1}}{d x_{1}}=0 \Rightarrow \frac{d y_{1}}{d x_{1}}=-\frac{\sqrt{y_{1}}}{\sqrt{x_{1}}}$........$.(i)$

Since the tangent to the given curve at $\left(x_{1}, y_{1}\right)$ is equally inclined to the axes.

$\therefore$ Slope of the tangent $\frac{d y_{1}}{d x}=\pm \tan \frac{\pi}{4}=\pm 1$

So, from eq. (i) we get

$-\frac{\sqrt{y_{1}}}{\sqrt{x_{1}}}=\pm 1$

So, from eq. $(i)$ we get

$-\frac{\sqrt{y_{1}}}{\sqrt{x_{1}}}=\pm 1$

On squaring on both the sides, we get

y1/x1 = 1 ⇒ y1 = x1

Now, putting the value of y1 in the given equation of the curve.

√x1 + √y1 = 4

√x1 + √x1 = 4 ⇒ 2√x1 = 4 ⇒ √x1 = 2 ⇒ x1 = 4

As y1 = x1

⇒ y1 = 4

Therefore, the required point is (4, 4).