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

Answer 1

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).