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

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(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

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

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(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:

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