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

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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?

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