Find the coordinates of the point on the curve $\sqrt{x} + \sqrt{y} = 4$ at which the tangent is equally inclined to the axes.
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → (4, 4)
Let $P(x_1,y_1)$ be a required point.
The given curve is $\sqrt{x} + \sqrt{y} = 4$ ...(i)
Differentiating (i) w.r.t. x, we get
$\frac{1}{2\sqrt{x}}+\frac{1}{2\sqrt{y}}.\frac{dy}{dx}=0⇒\frac{dy}{dx}=-\sqrt{\frac{y}{x}}$.
∴ The slope of tangent to curve (i) at $P(x_1,y_1) = -\sqrt{\frac{y_1}{x_1}}$
As the tangent to the curve is equally inclined to the coordinate axes, its slope = ±1.
∴ $-\sqrt{\frac{y_1}{x_1}}=1$ or $-\sqrt{\frac{y_1}{x_1}}=-1$ i.e. either $\sqrt{\frac{y_1}{x_1}}=-1$ or $\sqrt{\frac{y_1}{x_1}}=1$.
But $\sqrt{\frac{y_1}{x_1}}=-1$ is not possible (why?), so $\sqrt{\frac{y_1}{x_1}}=1⇒\sqrt{y_1}=\sqrt{x_1}$.
As the point $P(x_1,y_1)$ lies on the curve (i), so $\sqrt{x_1}+\sqrt{y_1} =4$
$\sqrt{x_1}+\sqrt{x_1} = 4 ⇒ \sqrt{x_1} =2⇒x_1 = 4$.
When $x_1 = 4, \sqrt{y_1}= \sqrt{4} ⇒ y_1 = 4$.
Hence, the required point is (4, 4).