The equation of the normal to the curve xy + | vedclass

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

Question

(A) Given the curve $xy + 7 = 0$, we have $y = -\frac{7}{x}$.Differentiating with respect to $x$, we get $\frac{dy}{dx} = \frac{7}{x^2}$.The slope of

Vedclass · Accepted Answer

$A > 0, B > 0$ or $A < 0, B < 0$

Vedclass · Answer

(A) Given the curve $xy + 7 = 0$, we have $y = -\frac{7}{x}$.Differentiating with respect to $x$, we get $\frac{dy}{dx} = \frac{7}{x^2}$.The slope of the tangent at any point $(x_0, y_0)$ is $m_t = \frac{7}{x_0^2}$.The slope of the normal $m_n$ is $-\frac{1}{m_t} = -\frac{x_0^2}{7}$.The equation of the normal at $(x_0, y_0)$ is $y - y_0 = -\frac{x_0^2}{7}(x - x_0)$.Multiplying by $7$, we get $7y - 7y_0 = -x_0^2 x + x_0^3$.Rearranging terms: $x_0^2 x + 7y - (7y_0 + x_0^3) = 0$.Comparing this with $Ax + By + C = 0$, we identify $A = x_0^2$ and $B = 7$.Since $x_0^2 > 0$ for any $x_0 
eq 0$ and $7 > 0$, both $A$ and $B$ must have the same sign (positive).Thus, $A > 0$ and $B > 0$.

The equation of the normal to the curve $xy + 7 = 0$ is $Ax + By + C = 0$. Which of the following is true?

Similar Questions

For the curve $x = a(\cos \theta + \theta \sin \theta)$,$y = a(\sin \theta - \theta \cos \theta)$,which of the following is true for the normal at any point $\theta$?

Find the equation of the tangent line to the curve $y=x^{2}-2x+7$ which is parallel to the line $2x-y+9=0$.

The tangent to the curve $y = ax^2 + bx$ at $(2, -8)$ is parallel to the $x$-axis. Then:

The tangent to the curve $y = xe^{x^2}$ passing through the point $(1, e)$ also passes through the point

Let the quadratic curve passing through the point $(-1, 0)$ and touching the line $y = x$ at $(1, 1)$ be $y = f(x)$. Then the $x$-intercept of the normal to the curve at the point $(\alpha, \alpha + 1)$ in the first quadrant is $..........$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module