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
- A \((4,4)\)
- B \((1,1)\)
- C \((9,9)\)
- D \((6,6)\)
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, \(\sqrt{x}+\sqrt{y}=6 \Rightarrow \sqrt{x}+\sqrt{x}=6 \Rightarrow 2 \sqrt{x}=6\)
\[
\begin{aligned}
\Rightarrow & \sqrt{x}=3 \Rightarrow x=9 \\
\therefore & x=9
\end{aligned}
\]
\[
\sqrt{x}+\sqrt{y}=6
\]
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, \(\sqrt{x}+\sqrt{y}=6 \Rightarrow \sqrt{x}+\sqrt{x}=6 \Rightarrow 2 \sqrt{x}=6\)
\[
\begin{aligned}
\Rightarrow & \sqrt{x}=3 \Rightarrow x=9 \\
\therefore & x=9
\end{aligned}
\]
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