What is the sum of the intercepts on the axes | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) Given the curve $\sqrt{x} + \sqrt{y} = 4$. Differentiating with respect to $x$,we get: $\frac{1}{2\sqrt{x}} + \frac{1}{2\sqrt{y}} \frac{dy}{dx} =

Vedclass · Accepted Answer

$16$

Vedclass · Answer

(D) Given the curve $\sqrt{x} + \sqrt{y} = 4$. Differentiating with respect to $x$,we get: $\frac{1}{2\sqrt{x}} + \frac{1}{2\sqrt{y}} \frac{dy}{dx} = 0$. $\frac{dy}{dx} = -\sqrt{\frac{y}{x}}$. At the point $(4, 4)$,the slope $m = -\sqrt{\frac{4}{4}} = -1$. The equation of the tangent at $(4, 4)$ is $y - 4 = -1(x - 4)$,which simplifies to $y - 4 = -x + 4$,or $x + y = 8$. To find the intercepts,we write the equation in intercept form: $\frac{x}{8} + \frac{y}{8} = 1$. The $x$-intercept is $8$ and the $y$-intercept is $8$. The sum of the intercepts is $8 + 8 = 16$.

What is the sum of the intercepts on the axes of the tangent to the curve $\sqrt{x} + \sqrt{y} = 4$ at the point $(4, 4)$?

Similar Questions

At what point does the tangent to the curve $y = e^{2x}$ at the point $(0, 1)$ meet the $x$-axis?

The slope of the normal to the curve $y=\frac{x}{x^2+1}$ at $x=-4$ is

Slope of the normal to the curve $y = x^{2} - \frac{1}{x^{2}}$ at the point $(-1, 0)$ is:

If the tangent to the curve $x^{2/3} + y^{2/3} = a^{2/3}$ meets the $X$-axis at $A$ and $Y$-axis at $B$,then $AB =$

An angle between the curves $x^2-y^2=4$ and $x^2+y^2=4 \sqrt{2}$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module