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  • Find the co-ordinates of the point on the curve root x +root y =4 at which tangent is equally inclined to the axes.

Find the co-ordinates of the point on the curve \sqrt x +\sqrt y =4 at which the tangent is equally inclined to the axes.

Answers (1)

Given: curve \sqrt x +\sqrt y =4 

To find: point coordinates on which tangent is equally inclined to the axis on the curve

Explanation: given \sqrt x +\sqrt y =4

After differentiating with respect to,

\frac{\mathrm{d}(\sqrt{\mathrm{x}}+\sqrt{\mathrm{y}})}{\mathrm{d} \mathrm{x}}=\frac{\mathrm{d}(4)}{\mathrm{dx}}$
Now using the sum rule of differentiation

\\ \frac{\mathrm{d}(\sqrt{\mathrm{x}})}{\mathrm{d} \mathrm{x}}+\frac{\mathrm{d}(\sqrt{\mathrm{y}})}{\mathrm{d} \mathrm{x}}=0$ \\$\frac{\mathrm{d}\left(\mathrm{x}^{\frac{1}{2}}\right)}{\mathrm{dx}}+\frac{\mathrm{d}\left(\mathrm{y}^{\frac{1}{2}}\right)}{\mathrm{d} \mathrm{x}}=0$
Then differentiating the equation, we get

\\ \Rightarrow \frac{1}{2} x^{\frac{1}{2}-1}+\frac{1}{2} y^{\frac{1}{2}-1} \frac{d(y)}{d x}=0 \\ \Rightarrow \frac{1}{2} x^{\frac{-1}{2}}+\frac{1}{2} y^{\frac{-1}{2}} \frac{d(y)}{d x}=0 \\ \Rightarrow \frac{1}{2} x^{\frac{-1}{2}}=-\frac{1}{2} y^{\frac{-1}{2}} \frac{d(y)}{d x} \\ \Rightarrow \frac{d y}{d x}=-\frac{x^{\frac{-1}{2}}}{y^{\frac{-1}{2}}} \\ \Rightarrow \frac{d y}{d x}=-\sqrt{\frac{y}{x}} \ldots . . \text { (i) }

The given curve has this tangent

As mentioned in the question tangent is equally inclined to the axis,

\\ \frac{d y}{d x}=\pm 1 \\\\ -\sqrt{\frac{y}{x}}=\pm 1 \\\\ \frac{y}{x}=1 \Rightarrow y=x

Substituting values in the curve equation from equation (ii)

\\$\sqrt y+\sqrt y=4$ \\$2 \sqrt y=4$ \\$ \sqrt y=2$ \\$\Rightarrow y=4$

When y = 4, then x = 4 from equation (ii)

Show the points on the curve $\sqrt{x}+\sqrt {y}=4$  at which the tangent equally inclined to the axis has the coordinates (4,4).

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