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

Answer: (B)

(9,9)

Answer 2

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).