If the line ax + by + c = 0 is a normal to th | vedclass

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

Question

(B) Given the curve $xy = 1$,we have $y = \frac{1}{x}$.Taking the derivative with respect to $x$,we get $\frac{dy}{dx} = -\frac{1}{x^2}$.The slope of

Vedclass · Accepted Answer

$a > 0, b < 0$

Vedclass · Answer

(B) Given the curve $xy = 1$,we have $y = \frac{1}{x}$.Taking the derivative with respect to $x$,we get $\frac{dy}{dx} = -\frac{1}{x^2}$.The slope of the tangent at any point $(x, y)$ is $-\frac{1}{x^2}$.The slope of the normal is the negative reciprocal of the tangent slope,which is $x^2$.The slope of the line $ax + by + c = 0$ is $-\frac{a}{b}$.Since the line is a normal to the curve,we equate the slopes: $x^2 = -\frac{a}{b}$.Since $x^2$ is always non-negative for real $x$,we must have $-\frac{a}{b} > 0$,which implies $\frac{a}{b} This condition holds if $a$ and $b$ have opposite signs,i.e.,$(a > 0, b  0)$.

If the line $ax + by + c = 0$ is a normal to the curve $xy = 1$,then

Similar Questions

Co-ordinates of a point on the curve $y = x \log x$ at which the normal is parallel to the line $2x - 2y = 3$ are

The intercepts on the $x$-axis made by tangents to the curve $y = \int_{0}^{x} |t| dt, x \in R$ which are parallel to the line $y = 2x$ are equal to:

If the curves $\frac{x^2}{a^2} + \frac{y^2}{4} = 1$ and $y^3 = 16x$ intersect at right angles,then $a^2 = \dots$

Find the equation of the tangent to the curve $y = be^{-x/a}$ at the point where it crosses the $y$-axis.

For the curve ${y^n} = {a^{n - 1}}x$,the subnormal at any point is constant. The value of $n$ must be

Vedclass Test Series

Exam Paper Generator

Online Exam Module