If the line ax + by + c = 0 is a tangent to t | vedclass

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

Question

(C) The equation of the curve is $xy = 4$, which can be written as $y = \frac{4}{x}$.Taking the derivative with respect to $x$, we get $\frac{dy}{dx}

Vedclass · Accepted Answer

$a < 0, b < 0$

Vedclass · Answer

(C) The equation of the curve is $xy = 4$, which can be written as $y = \frac{4}{x}$.Taking the derivative with respect to $x$, we get $\frac{dy}{dx} = -\frac{4}{x^2}$.The equation of the line is $ax + by + c = 0$, which can be rewritten as $y = -\frac{a}{b}x - \frac{c}{b}$.The slope of this line is $m = -\frac{a}{b}$.Since the line is a tangent to the curve, the slope of the tangent at any point $(x, y)$ on the curve must equal the slope of the line:$-\frac{4}{x^2} = -\frac{a}{b} \implies \frac{a}{b} = \frac{4}{x^2}$.Since $x^2 > 0$ for all $x 
eq 0$, it follows that $\frac{a}{b} > 0$.This implies that $a$ and $b$ must have the same sign. Looking at the options, the condition $a  0$.

If the line $ax + by + c = 0$ is a tangent to the curve $xy = 4$, then

Similar Questions

If the tangent to the curve $y = x^3 + ax - b$ at the point $(1, -5)$ is perpendicular to the line $-x + y + 4 = 0$,then which one of the following points lies on the curve?

Find all points on the curve $y = 4x^3 - 2x^5$ at which the tangent passes through the origin.

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

For the curve $x = a(\theta + \sin \theta)$ and $y = a(1 - \cos \theta)$,the lengths of the tangent and the sub-tangent at point $\theta$ are respectively:

$V$ is the set of points on the curve $y^3 - 3xy + 2 = 0$ where the tangent is vertical. Then $V = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module