AP DSC School Assistant Math Model Test 2

Concept:The slopes of parallel lines are equal. For a curve, the slope of the tangent at a point is given by the derivative $\frac{dy}{dx}$.Explanation:1. The given hyperbola is $x^2 - y^2 = 3$. Differentiating both sides with respect to $x$: $2x - 2y \frac{dy}{dx} = 0 \Rightarrow \frac{dy}{dx} = \frac{x}{y}$. So the slope of the tangent at any point $(x,y)$ on the hyperbola is $m_1 = \frac{x}{y}$.2. The given straight line is $2x + y + 8 = 0$. Its slope is found by rewriting in the form $y = mx + c$ or by differentiating. Differentiating: $2 + \frac{dy}{dx} = 0 \Rightarrow \frac{dy}{dx} = -2$. Hence the slope of the line is $m_2 = -2$.3. For the tangent to be parallel to the line, their slopes must be equal: $m_1 = m_2 \Rightarrow \frac{x}{y} = -2 \Rightarrow x = -2y$. …(i)4. Substitute $x = -2y$ into the hyperbola equation: $(-2y)^2 - y^2 = 3 \Rightarrow 4y^2 - y^2 = 3 \Rightarrow 3y^2 = 3 \Rightarrow y^2 = 1 \Rightarrow y = \pm 1$.5. Using (i): - If $y = 1$, then $x = -2(1) = -2$, giving point $(-2,1)$. - If $y = -1$, then $x = -2(-1) = 2$, giving point $(2,-1)$.Answer:The required points are $(2, -1)$ and $(-2, 1)$, which corresponds to option B.