A(1, -3) and B(4, 3) are two points on the cu | vedclass

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

Question

(B) The slope of the chord $AB$ is given by $m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{3 - (-3)}{4 - 1} = \frac{6}{3} = 2$. Since the tangent is paralle

Vedclass · Accepted Answer

$(2, 0), (-2, 0)$

Vedclass · Answer

(B) The slope of the chord $AB$ is given by $m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{3 - (-3)}{4 - 1} = \frac{6}{3} = 2$. Since the tangent is parallel to the chord $AB$,the slope of the tangent must also be $2$. Given the curve $y = x - \frac{4}{x}$,we find the derivative $\frac{dy}{dx} = 1 + \frac{4}{x^2}$. Setting the derivative equal to the slope of the chord: $1 + \frac{4}{x^2} = 2$. This simplifies to $\frac{4}{x^2} = 1$,which implies $x^2 = 4$,so $x = \pm 2$. For $x = 2$,$y = 2 - \frac{4}{2} = 0$. For $x = -2$,$y = -2 - \frac{4}{-2} = -2 + 2 = 0$. Thus,the required points are $(2, 0)$ and $(-2, 0)$.

$A(1, -3)$ and $B(4, 3)$ are two points on the curve $y = x - \frac{4}{x}$. The points on the curve,the tangents at which are parallel to the chord $AB$,are

Similar Questions

The curve $y=x^3-2x^2+3x-4$ intersects the horizontal line $y=-2$ at the point $P(h, k)$. If the tangent drawn to this curve at $P$ meets the $X$-axis at $(x_1, y_1)$,then $x_1=$

The normal to the curve $y=f(x)$ at the point $(3,4)$ makes an angle $\frac{3 \pi}{4}$ with the positive $X$-axis. Then $f^{\prime}(3)$ is equal to:

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

If the equation of the normal to the curve $x^3 - y^2 = 0$ at the point $(m^2, -m^3)$ is $y = 3mx - 4m^3$,then $m^2 = \dots\dots$.

The sum of the intercepts on the axes cut off by the tangent to the curve $x^{2/3} + y^{2/3} = 2$ at the point $(1, 1)$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module