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KCET · Maths · Application of Derivatives

The coordinates of the point on the \(\sqrt{x}+\sqrt{y}=6\) at which the tangent is equally inclined to the axes is

  1. A \((4,4)\)
  2. B \((1,1)\)
  3. C \((9,9)\)
  4. D \((6,6)\)
Verified Solution

Answer & Solution

Correct Answer

(C) \((9,9)\)

Step-by-step Solution

Detailed explanation

Given, \(\sqrt{x}+\sqrt{y}=6\)
\[
\sqrt{x}+\sqrt{y}=6
\]
Differentiating w.r.t. \(x\), we get
\[
\frac{1}{2 \sqrt{x}}+\frac{1}{2 \sqrt{y}} \frac{d y}{d x}=0 \Rightarrow \frac{d y}{d x}=-\frac{\sqrt{y}}{\sqrt{x}}
\]
Since, the tangent is equally inclined to the axes.
\[
\begin{aligned}
& \frac{d y}{d x}= \pm 1 \Rightarrow-\frac{\sqrt{y}}{\sqrt{x}}= \pm 1 \\
& \Rightarrow \quad \sqrt{y}= \pm \sqrt{x} \Rightarrow y=x \\
&
\end{aligned}
\]
Therefore, \(\sqrt{x}+\sqrt{y}=6 \Rightarrow \sqrt{x}+\sqrt{x}=6 \Rightarrow 2 \sqrt{x}=6\)
\[
\begin{aligned}
\Rightarrow & \sqrt{x}=3 \Rightarrow x=9 \\
\therefore & x=9
\end{aligned}
\]