Find the Tangent Given the Point (-2, square root of 2)

Problem

x2+y2=6,(−2,√(,2))

Solution

1. Identify the equation of the circle and the given point ((x_1),(y_1))=(−2,√(,2))

2. Differentiate the equation x2+y2=6 implicitly with respect to x to find the slope of the tangent line.

d(x2)/d(x)+d(y2)/d(x)=d(6)/d(x)

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

3. Solve for the derivative d(y)/d(x) to determine the general slope formula.

2*yd(y)/d(x)=−2*x

d(y)/d(x)=−x/y

4. Substitute the coordinates of the point (−2,√(,2)) into the derivative to find the specific slope m

m=−(−2)/√(,2)

m=2/√(,2)

m=√(,2)

5. Apply the point-slope formula y−(y_1)=m*(x−(x_1)) using the point (−2,√(,2)) and the slope m=√(,2)

y−√(,2)=√(,2)*(x−(−2))

y−√(,2)=√(,2)*(x+2)

6. Simplify the equation into slope-intercept form.

y−√(,2)=√(,2)*x+2√(,2)

y=√(,2)*x+3√(,2)

Final Answer

y=√(,2)*x+3√(,2)

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