The lines y = -frac{3}{2} x and y = -frac{2}{ | vedclass

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(C) Given the curve $3x^2 + 4xy + 5y^2 - 4 = 0$. Differentiating with respect to $x$,we get $6x + 4y + 4x\frac{dy}{dx} + 10y\frac{dy}{dx} = 0$.Thus,$\

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are perpendicular to each other

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(C) Given the curve $3x^2 + 4xy + 5y^2 - 4 = 0$. Differentiating with respect to $x$,we get $6x + 4y + 4x\frac{dy}{dx} + 10y\frac{dy}{dx} = 0$.Thus,$\frac{dy}{dx} = -\frac{6x + 4y}{4x + 10y} = -\frac{3x + 2y}{2x + 5y}$.For point $P$ on the line $y = -\frac{3}{2}x$,we have $2y = -3x$,so $3x + 2y = 0$. Substituting this into the derivative,$\left. \frac{dy}{dx} \right|_P = 0$.For point $Q$ on the line $y = -\frac{2}{5}x$,we have $5y = -2x$,so $2x + 5y = 0$. Substituting this into the derivative,$\left. \frac{dy}{dx} \right|_Q = \infty$.Since the slope at $P$ is $0$ (horizontal tangent) and the slope at $Q$ is $\infty$ (vertical tangent),the tangents are perpendicular to each other.

The lines $y = -\frac{3}{2} x$ and $y = -\frac{2}{5} x$ intersect the curve $3x^2 + 4xy + 5y^2 - 4 = 0$ at the points $P$ and $Q$ respectively. The tangents drawn to the curve at $P$ and $Q$:

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