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

The curve $y=ax^3+bx^2+cx+5$ touches the $X$-axis at $(-2,0)$ and cuts the $Y$-axis at a point $Q$ where its gradient is $3$. Then the values of $a, b, c$ respectively are:

The equation of the tangent to the curve $y = \sin x$ at the point $(\pi, 0)$ is $......$.

For the curve $x = a(\cos \theta + \theta \sin \theta)$,$y = a(\sin \theta - \theta \cos \theta)$,which of the following is true for the normal at any point $\theta$?

Find the point on the curve $y=x^{3}-11x+5$ at which the tangent is $y=x-11$.

Let the area enclosed by the $x$-axis,and the tangent and normal drawn to the curve $4x^{3}-3xy^{2}+6x^{2}-5xy-8y^{2}+9x+14=0$ at the point $(-2,3)$ be $A$. Then $8A$ is equal to $.......$

Vedclass Test Series

Exam Paper Generator

Online Exam Module