AP Calculus ABmediummcq1 pt

The curve $x^2+4y^2=8$ passes through the point $(2,1)$ . An equation of the tangent line to the curve at this point is

y-1=\dfrac{1}{2}(x-2)

y-1=-2(x-2)

y-1=-\dfrac{1}{2}(x-2)

y-1=-(x-2)

Explanation

Core Concept

Implicit differentiation gives $2x+8y\frac{dy}{dx}=0$ , so $\frac{dy}{dx}=-\frac{x}{4y}=-\frac{1}{2}$ at $(2,1)$ . Point-slope form gives $y-1=-\frac{1}{2}(x-2)$ .

Correct Answer

C $y-1=-\dfrac{1}{2}(x-2)$

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The curve x2+4y2=8x^2+4y^2=8x2+4y2=8 passes through the point (2,1)(2,1)(2,1). An equation of the tangent line to the curve at this point is

Explanation

Core Concept

Correct Answer

More Unit 3: Differentiation: Composite, Implicit, and Inverse Functions practice questions

The curve $x^2+4y^2=8$ passes through the point $(2,1)$ . An equation of the tangent line to the curve at this point is