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 \(\sqrt{y} = 6 - \sqrt{x}\) and then square both sides:
\(y = (6 - \sqrt{x})^2\)
\(y = 36 - 12\sqrt{x} + x\)
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, we have \(36 - 12\sqrt{x} = 0\)
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, we have \(36 - 12\sqrt{x} = -2x\)
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).