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Question
Find the co-ordinates of the point on the curve `sqrt(x) + sqrt(y)` = 4 at which tangent is equally inclined to the axes
Solution
Equation of curve is given by `sqrt(x) + sqrt(y)` = 4
Let (x1, y1) be the required point on the curve
∴ `sqrt(x)_1 + sqrt(y)_1` = 4
Differentiating both sides w.r.t. x1, we get
`"d"/("dx"_1) sqrt(x_1) + "d"/("dx"_1) sqrt(y_1) = "d"/("dx"_1) (4)`
⇒ `1/(2sqrt(x_1)) + 1/(2sqrt(y_1)) * ("d"y_1)/("dx"_1)` = 0
⇒ `1/sqrt(x_1) + 1/sqrt(y_1) * ("dy"_1)/("dx"_1)` = 0
⇒ `("dy"_1)/("d"x_1) = - sqrt(y_1)/sqrt(x_1)` .....(i)
Since the tangent to the given curve at (x1, y1) is equally inclined to the axes.
∴ Slope of the tangent `("dy"_1)/("dx"_1) = +- tan pi/4` = ±1
So, from equation (i) we get
`- sqrt(y_1)/sqrt(x_1)` = ±1
Squaring both sides, we get
`(y_1)/(x_1)` = 1
⇒ y1 = x1
Putting the value of y1 in the given equation of the curve.
`sqrt(x_1) + sqrt(y_1)` = 4
⇒ `sqrt(x_1) + sqrt(x_1)` = 4
⇒ `2sqrt(x_1)` = 4
⇒ `sqrt(x_1)` = 2
⇒ x1 = 4
Since y1 = x1
∴ y1 = 4
Hence, the required point is (4, 4).
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