If (a^2-1) x+a y+(3-a)=0 is a normal to the c | vedclass

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

Question

(B) Given the curve $x y=1$,we have $y = \frac{1}{x}$.Calculating the derivative: $\frac{d y}{d x} = -\frac{1}{x^2}$.The slope of the tangent at any p

Vedclass · Accepted Answer

$(-\infty,-1] \cup(0,1]$

Vedclass · Answer

(B) Given the curve $x y=1$,we have $y = \frac{1}{x}$.Calculating the derivative: $\frac{d y}{d x} = -\frac{1}{x^2}$.The slope of the tangent at any point $(x, y)$ is $m_t = -\frac{1}{x^2}$.The slope of the normal $m_n$ is the negative reciprocal of the tangent slope: $m_n = -\frac{1}{m_t} = x^2$.Since $x^2 \geq 0$ for all real $x$,the slope of the normal must be non-negative $(m_n \geq 0)$.The given equation of the normal is $(a^2-1) x+a y+(3-a)=0$,which can be rewritten as $a y = -(a^2-1) x - (3-a)$,or $y = -\frac{a^2-1}{a} x - \frac{3-a}{a}$.The slope of this line is $m = -\frac{a^2-1}{a} = \frac{1-a^2}{a}$.Since $m \geq 0$,we have $\frac{1-a^2}{a} \geq 0$.Multiplying by $-1$ reverses the inequality: $\frac{a^2-1}{a} \leq 0$,which is $\frac{(a-1)(a+1)}{a} \leq 0$.Using the sign scheme (Wavy Curve Method),the values of $a$ satisfying this are $a \in (-\infty, -1] \cup (0, 1]$.

If $(a^2-1) x+a y+(3-a)=0$ is a normal to the curve $x y=1$,then the interval in which '$a$' lies is

Similar Questions

If $\theta$ is an angle between the curves $x^2+4y=0$ and $xy=2$,then $\tan \theta=$

If the angle between the curves $y=e^{2(1+x)-4}$ and $x^2 y=1$ at the point $(1,1)$ is $\theta$,then $|\sin \theta|+|\cos \theta|=$

If $y = 4x - 6$ is a tangent to the curve $y^2 = ax^4 + b$ at $(3, 6)$,then the values of $a$ and $b$ are:

If an equation of a tangent to the curve $y = \cos(x + y)$,where $-1 - \pi \le x \le 1 + \pi$,is $x + 2y = k$,then $k$ is equal to

At which point on the curve $y = x^2 + 3x$ should the tangent be drawn so that it passes through the point $(0, -9)$?

Vedclass Test Series

Exam Paper Generator

Online Exam Module