The sum of the intercepts made by a tangent d | vedclass

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

Question

(D) Given the curve equation: $\left(\frac{x}{a}\right)^n + \left(\frac{y}{b}\right)^n = 2$. Taking the derivative with respect to $x$: $\frac{n}{a} \

Vedclass · Accepted Answer

$2(a+b)$

Vedclass · Answer

(D) Given the curve equation: $\left(\frac{x}{a}\right)^n + \left(\frac{y}{b}\right)^n = 2$. Taking the derivative with respect to $x$: $\frac{n}{a} \left(\frac{x}{a}\right)^{n-1} + \frac{n}{b} \left(\frac{y}{b}\right)^{n-1} \frac{dy}{dx} = 0$. At the point $(a, b)$,the slope $m$ is: $m = \left(\frac{dy}{dx}\right)_{(a,b)} = -\frac{b}{a} \left(\frac{a/a}{b/b}\right)^{n-1} = -\frac{b}{a}$. The equation of the tangent at $(a, b)$ is: $y - b = -\frac{b}{a}(x - a) \Rightarrow ay - ab = -bx + ab \Rightarrow bx + ay = 2ab$. Dividing by $2ab$,we get: $\frac{x}{2a} + \frac{y}{2b} = 1$. The intercepts on the axes are $2a$ and $2b$. The sum of the intercepts is $2a + 2b = 2(a+b)$.

The sum of the intercepts made by a tangent drawn to the curve $\left(\frac{x}{a}\right)^n + \left(\frac{y}{b}\right)^n = 2$ at the point $(a, b)$ on the coordinate axes is:

Similar Questions

If the locus of the points on the curve $x^3 y^2+\frac{x^2}{y}=5$ at which the tangent is parallel to the $X$-axis is $f(x, y)=0$,then the point that lies on this curve $f(x, y)=0$ is

The length of the subtangent,ordinate,and the subnormal are in

The coordinates of the point on the curve $y=x^2-3x+2$ where the tangent is perpendicular to the straight line $y=x$ are

If the normal to the curve $y = f(x)$ at the point $(4, 6)$ makes an angle $\frac{2\pi}{3}$ with the positive $x$-axis in the anticlockwise direction,then $f'(4)$ is:

The angle of intersection of the curves $y = x^2$ and $x = y^2$ at $(1, 1)$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module