Question

The co-ordinates 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. (9,9)
• C. (1,1)
• D. (6,6)

Answer 1

The Correct Option is B

Given the equation:

\[ \sqrt{x} + \sqrt{y} = 6 \]

We can rewrite it as:

\[ \sqrt{y} = 6 - \sqrt{x} \]

Now, we square both sides:

\[ y = (6 - \sqrt{x})^2 \]

Expanding the right-hand side:

\[ y = 36 - 12\sqrt{x} + x \]

To find the point of interest, we need to find the intersection of the curve with the slopes of +1 and -1.

For the slope of +1, we have \(y = x\). Substituting this into the equation of the curve, we get:

\[ 36 - 12\sqrt{x} + x = x \]

Simplifying:

\[ 36 - 12\sqrt{x} = 0 \]

Rearranging, we find:

\[ \sqrt{x} = 3 \]

Squaring both sides, we get \(x = 9\). Substituting this value back into the equation \(y = x\), we find \(y = 9\).

So, for the slope of +1, the point of interest is \((9, 9)\).

For the slope of -1, we have \(y = -x\). Substituting this into the equation of the curve, we get:

\[ 36 - 12\sqrt{x} + x = -x \]

Simplifying:

\[ 36 - 12\sqrt{x} = -2x \]

Rearranging, we find:

\[ \sqrt{x} = -3 \]

However, the square root is defined only for non-negative values, so there is no solution for \(\sqrt{x} = -3\).

Therefore, there is no point of interest with a slope of -1.

Hence, the point on the curve \(\sqrt{x} + \sqrt{y} = 6\) at which the tangent is equally inclined to the axes is \((9, 9)\), which corresponds to option (B) \((9, 9)\).

Answer 2

We are given: \[ \sqrt{x} + \sqrt{y} = 6 \] Let: \[ u = \sqrt{x},\quad v = \sqrt{y} \Rightarrow u + v = 6 \Rightarrow v = 6 - u \] So: \[ y = v^2 = (6 - u)^2 = 36 - 12u + u^2 \quad \text{and} \quad x = u^2 \] 

Step 1: Differentiate the original equation
\[ \sqrt{x} + \sqrt{y} = 6 \Rightarrow \frac{1}{2\sqrt{x}} + \frac{1}{2\sqrt{y}} \cdot \frac{dy}{dx} = 0 \] Multiply both sides by 2: \[ \frac{1}{\sqrt{x}} + \frac{1}{\sqrt{y}} \cdot \frac{dy}{dx} = 0 \Rightarrow \frac{dy}{dx} = -\frac{\sqrt{y}}{\sqrt{x}} \] 

Step 2: Tangent equally inclined to axes
This means the slope is either \(1\) or \(-1\) So: \[ -\frac{\sqrt{y}}{\sqrt{x}} = 1 \text{ or } -1 \Rightarrow \frac{\sqrt{y}}{\sqrt{x}} = -1 \text{ or } 1 \Rightarrow \sqrt{y} = \sqrt{x} \Rightarrow x = y \] Substitute into original: \[ \sqrt{x} + \sqrt{x} = 6 \Rightarrow 2\sqrt{x} = 6 \Rightarrow \sqrt{x} = 3 \Rightarrow x = 9 \Rightarrow y = 9 \] 

Final answer: (9, 9)

Answer 3

We are given the equation \(\sqrt{x} + \sqrt{y} = 6\). We need to find the coordinates of the point where the tangent is equally inclined to the axes.

If the tangent is equally inclined to the axes, its slope (dy/dx) must be either 1 or -1. Since the sum of square roots is a decreasing relation, let us take m = -1.

First, differentiate the given equation implicitly with respect to x:

\(\frac{d}{dx}(\sqrt{x} + \sqrt{y}) = \frac{d}{dx}(6)\)

\(\frac{1}{2\sqrt{x}} + \frac{1}{2\sqrt{y}}\frac{dy}{dx} = 0\)

\(\frac{1}{2\sqrt{y}}\frac{dy}{dx} = -\frac{1}{2\sqrt{x}}\)

\(\frac{dy}{dx} = -\frac{\sqrt{y}}{\sqrt{x}}\)

We want the slope (dy/dx) to be -1, so:

-\(\frac{\sqrt{y}}{\sqrt{x}}\) = -1

\(\sqrt{y} = \sqrt{x}\)

y = x

Substitute y = x into the original equation:

\(\sqrt{x} + \sqrt{x} = 6\)

2\(\sqrt{x}\) = 6

\(\sqrt{x}\) = 3

x = 9

Since y = x, then y = 9.

Therefore, the coordinates of the point are (9, 9).

Answer: (9,9)