The tangent to the curve xy = 25 at any point | vedclass

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

Question

(A) Given curve is $xy = 25$. Let the point of tangency be $P(x_1, y_1)$. Since $P$ lies on the curve,$x_1 y_1 = 25$.Differentiating $xy = 25$ with re

Vedclass · Accepted Answer

$50$ sq units

Vedclass · Answer

(A) Given curve is $xy = 25$. Let the point of tangency be $P(x_1, y_1)$. Since $P$ lies on the curve,$x_1 y_1 = 25$.Differentiating $xy = 25$ with respect to $x$,we get $y + x \frac{dy}{dx} = 0$,so $\frac{dy}{dx} = -\frac{y}{x}$.At $(x_1, y_1)$,the slope of the tangent is $m = -\frac{y_1}{x_1}$.The equation of the tangent is $y - y_1 = -\frac{y_1}{x_1}(x - x_1)$.Multiplying by $x_1$,we get $x_1 y - x_1 y_1 = -y_1 x + x_1 y_1$,which simplifies to $x_1 y + y_1 x = 2 x_1 y_1$.Dividing by $2 x_1 y_1$,we get $\frac{x}{2 x_1} + \frac{y}{2 y_1} = 1$.The tangent cuts the $x$-axis at $A(2 x_1, 0)$ and the $y$-axis at $B(0, 2 y_1)$.The area of $	riangle OAB = \frac{1}{2} 	imes 	ext{base} 	imes 	ext{height} = \frac{1}{2} 	imes (2 x_1) 	imes (2 y_1) = 2 x_1 y_1$.Since $x_1 y_1 = 25$,the area is $2(25) = 50$ sq units.

The tangent to the curve $xy = 25$ at any point on it cuts the coordinate axes at $A$ and $B$. Then the area of the $\triangle OAB$ is

Similar Questions

Show that the normal at any point $\theta$ to the curve $x=a \cos \theta+a \theta \sin \theta, y=a \sin \theta-a \theta \cos \theta$ is at a constant distance from the origin.

The point on the curve $y=x^3$,at which the tangent to the curve is parallel to the $X$-axis,is

Let $f(x) = \begin{cases} -x^2 & \text{for } x < 0 \\ x^2 + 8 & \text{for } x \ge 0 \end{cases}$. Then the $x$-intercept of the line that is tangent to the graph of $f(x)$ is

The normal to the curve $x = a(\cos \theta + \theta \sin \theta )$ and $y = a(\sin \theta - \theta \cos \theta )$ at any $\theta$ is such that:

The equation of a normal to the curve $x=4 \sec \theta$ and $y=4 \tan^2 \theta$ at $\theta=\frac{\pi}{4}$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module