Find the Tangent Line at the Point 2x^2+y^2=17 , (2,3)

Problem

2*x2+y2=17,(2,3)

Solution

1. Differentiate implicitly with respect to x to find the expression for the slope of the tangent line.

(d(2)*x2)/d(x)+d(y2)/d(x)=d(17)/d(x)

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

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

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 (2,3) into the derivative.

m=−(2*(2))/3

m=−4/3

4. Apply the point-slope form y−(y_1)=m*(x−(x_1)) using the point (2,3) and the slope m=−4/3

y−3=−4/3*(x−2)

5. Simplify the equation into slope-intercept form y=m*x+b

y−3=−4/3*x+8/3

y=−4/3*x+8/3+9/3

y=−4/3*x+17/3

Final Answer

y=−4/3*x+17/3

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