The sum of intercepts on coordinate axes made | vedclass

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

Question

(A) Given the curve $\sqrt{x}+\sqrt{y}=\sqrt{a}$.Differentiating with respect to $x$,we get $\frac{1}{2\sqrt{x}}+\frac{1}{2\sqrt{y}}\frac{dy}{dx}=0$.T

Vedclass · Accepted Answer

$a$

Vedclass · Answer

(A) Given the curve $\sqrt{x}+\sqrt{y}=\sqrt{a}$.Differentiating with respect to $x$,we get $\frac{1}{2\sqrt{x}}+\frac{1}{2\sqrt{y}}\frac{dy}{dx}=0$.Thus,the slope of the tangent is $\frac{dy}{dx}=-\sqrt{\frac{y}{x}}$.Let the point of tangency be $(x_1, y_1)$. The equation of the tangent is $(y-y_1)=-\sqrt{\frac{y_1}{x_1}}(x-x_1)$.Multiplying by $\sqrt{x_1}$,we get $\sqrt{x_1}y - \sqrt{x_1}y_1 = -\sqrt{y_1}x + \sqrt{y_1}x_1$.Rearranging gives $\sqrt{y_1}x + \sqrt{x_1}y = \sqrt{x_1}y_1 + \sqrt{y_1}x_1 = \sqrt{x_1y_1}(\sqrt{x_1}+\sqrt{y_1})$.Since $\sqrt{x_1}+\sqrt{y_1}=\sqrt{a}$,the equation becomes $\sqrt{y_1}x + \sqrt{x_1}y = \sqrt{x_1y_1a}$.Dividing by $\sqrt{x_1y_1a}$,we get $\frac{x}{\sqrt{x_1a}} + \frac{y}{\sqrt{y_1a}} = 1$.The $x$-intercept is $\sqrt{x_1a}$ and the $y$-intercept is $\sqrt{y_1a}$.The sum of intercepts is $\sqrt{a}(\sqrt{x_1}+\sqrt{y_1}) = \sqrt{a} 	imes \sqrt{a} = a$.

The sum of intercepts on coordinate axes made by the tangent to the curve $\sqrt{x}+\sqrt{y}=\sqrt{a}$ is

Similar Questions

If the normal drawn at a point $P$ on the curve $3y = 6x - 5x^3$ passes through $(0,0)$,then the positive integral value of the abscissa of the point $P$ is

The angle of intersection of the curves $r = \sin \theta + \cos \theta$ and $r = 2 \sin \theta$ is equal to:

The length of the subtangent at any point $(x_1, y_1)$ on the curve $y=5^x$ is

If the tangent to the curve $y^{2} = x^{3}$ at $(m^{2}, m^{3})$ is also a normal to the curve at $(M^{2}, M^{3})$, then the value of $mM$ is

If the angle between the curves $y^2=4x$ and $y=e^{-x/2}$ is $\theta$,then $\operatorname{cosec}^2(\theta/2)=$

Vedclass Test Series

Exam Paper Generator

Online Exam Module