Find all points on the curve [tex]xy^2 - 3x^2y = 10[/tex] whose x-coordinate is 1. Then, write the equation for the tangent line at each of those points.
Answer (3)
xy²-3x²y=10 whose x-coordinate is 1 this part, just plug in the one for x and solve foe y (1)y²-3(1)²y=10 y²-3y=10 y²-3y-10=0 (y-5)(y+2)=0 y-5=0 & y+2=0 y=5, -2 when x =1, y=5 and -2 (1,5) & (1,-2) To find the tangent line, find the derivative and set to y' xy²-3x²y=10 x(2yy')+y²(1)-3x²(y')+y(-6x)=0 2xyy'+y²-3x²y'-6xy=0 y'(2xy-3x²)=6xy-y² y'=6xy-y²/(2xy-3x²) Plug in the two points from the first part (1,5) 6(1)(5)-(5)²/(2(1)(5)-3(1)²) 30-25/(10-3) =5/7 <-- slope of tangent line at (1,5) put the point and slope in point-slope form (y-5)=5/7(x-1) <-- your tangent equation, then you repeat the same for (1,-2) I hope you can try (1,-2) by yourself, good luck.
To find the points on the curve xy^2 - 3x^2y = 10 with an x-coordinate of 1, substitute x with 1 into the equation and solve for y. Differentiate the equation implicitly to find the slope of the tangent line, then use the point-slope formula to write the tangent line's equation.
The student's question asks to find all points on the curve with the equation xy^2 - 3x^2y = 10 whose x-coordinate is 1, and to write the equation for the tangent line at those points. When x = 1, we can substitute into the curve's equation and solve for y. Then, by differentiating the curve's equation implicitly with respect to x, we can find the slope of the tangent line at those points, which is ∂y/∂x. Lastly, we use the point-slope form of the line, y - y1 = m(x - x1), where (x1, y1) is the point of tangency and m is the slope obtained from the derivative, to write the equation of the tangent line.
There are two points on the curve x y 2 − 3 x 2 y = 10 where x = 1 , namely ( 1 , 5 ) and ( 1 , − 2 ) . The slopes of the tangent lines at these points are 7 5 and 7 16 respectively. The tangent line equations are derived using the point-slope form.
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