Find the Tangent Line at the Point 2x^2+y^2=41 , (-4,3)

Problem

2*x2+y2=41,(−4,3)

Solution

1. Differentiate implicitly with respect to x to find the slope of the curve at any point.

(d(2)*x2)/d(x)+d(y2)/d(x)=d(41)/d(x)

4*x+2*yd(y)/d(x)=0

2. Solve for the derivative d(y)/d(x) to isolate the expression for the slope.

2*yd(y)/d(x)=−4*x

d(y)/d(x)=−(4*x)/(2*y)

d(y)/d(x)=−(2*x)/y

3. Evaluate the slope m by substituting the given point (x,y)=(−4,3) into the derivative.

m=−(2*(−4))/3

m=8/3

4. Apply the point-slope formula y−(y_1)=m*(x−(x_1)) using the point (−4,3) and the slope m=8/3

y−3=8/3*(x−(−4))

y−3=8/3*(x+4)

5. Simplify to slope-intercept form by distributing the slope and adding 3 to both sides.

y−3=8/3*x+32/3

y=8/3*x+32/3+9/3

y=8/3*x+41/3

Final Answer

y=8/3*x+41/3

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