Let the tangent to a curve at point (x, y) in | vedclass

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

Question

(A) Let the point on the curve be $(x, y)$. The slope of the tangent is $m_1 = \frac{dy}{dx}$.The equation of the tangent is $Y - y = \frac{dy}{dx}(X

Vedclass · Accepted Answer

$y \frac{dy}{dx} - x \left(\frac{dy}{dx}\right)^2 = 1$

Vedclass · Answer

(A) Let the point on the curve be $(x, y)$. The slope of the tangent is $m_1 = \frac{dy}{dx}$.The equation of the tangent is $Y - y = \frac{dy}{dx}(X - x)$.To find the intersection with the $Y$-axis, set $X = 0$: $Y_P = y - x \frac{dy}{dx}$. So, $P = (0, y - x \frac{dy}{dx})$.The slope of the line perpendicular to the tangent is $m_2 = -\frac{1}{dy/dx} = -\frac{dx}{dy}$.This line passes through $P(0, y - x \frac{dy}{dx})$ and $(1, 0)$.The slope $m_2$ is $\frac{0 - (y - x \frac{dy}{dx})}{1 - 0} = -y + x \frac{dy}{dx}$.Equating the two expressions for $m_2$: $-\frac{dx}{dy} = -y + x \frac{dy}{dx}$.Multiply by $-\frac{dy}{dx}$: $1 = y \frac{dy}{dx} - x \left(\frac{dy}{dx}\right)^2$.Thus, $y \frac{dy}{dx} - x \left(\frac{dy}{dx}\right)^2 = 1$.

Let the tangent to a curve at point $(x, y)$ intersect the $Y$-axis at point $P$. $A$ line drawn through point $P$ is perpendicular to this tangent and passes through the point $(1, 0)$. The differential equation of the curve is:

Similar Questions

The general solution of the differential equation of all circles having center at $A(-1, 2)$ is $ . . . . . . $.

Find the differential equation for the family of curves given by $y = e^{x}(a \cos x + b \sin x)$ by eliminating the arbitrary constants $a$ and $b$.

The elimination of the arbitrary constants $A, B$ and $C$ from $y = A + Bx + C{e^{ - x}}$ leads to the differential equation:

The differential equation for all the straight lines which are at a unit distance from the origin is

The differential equation for the family of curves ${x^2} + {y^2} - 2ay = 0,$ where $a$ is an arbitrary constant,is

Vedclass Test Series

Exam Paper Generator

Online Exam Module