Answer :
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.
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.
