find the tangents of the following curves y^2 = 12x from (2, 7). the answers are 3x - y +1 = 0 and x- 2y + 12 = 0. helppp
calculus? first, use implicit differentiation on y^2 = 12 can you do that?
2y dy/dx = 12
dy/dx = 12/ 2y
ok, or dy/dx= 6/y
so dy/dx = 6/7
let a point on the curve be (x,y) the slope from that point to (2,7) is \[ m = \frac{y-7}{x-2} \] we want this slope to match dy/dx ( the derivative is the slope) \[ \frac{6}{y} = \frac{y-7}{x-2} \] we could solve that if we got rid of the x. the curve is y^2 = 12 x and so \[ x= \frac{y^2}{12} \] use that value for x in our equation \[ \frac{6}{y} = \frac{y-7}{\frac{y^2}{12}-2} \]
can you solve for y ?
sure, just wait sir :)
its decimal.. not whole number or i was just too stupid solving algebra..
what do you get after cross multiplying ?
6 (y^2 / 12 -2) = y (y-7) 6y^2 / 12 -12 = y^2 - 7y 6y^2 - y^2 = -7y + 12 1/2 y^2 - 7y -12 = 0..
am i right?? im sorry
6y^2 / 12 -12 = y^2 - 7y at this step you have \[ \frac{6}{12} y^2 - 12 = y^2 -7y \\ \frac{1}{2} y^2 - 12 = y^2 -7y \] I would multiply both sides by 2 to get rid of the fraction \[ y^2 -24= 2y^2 -14y \] can you finish ?
its still the same.. 15. 54 and -1.5
noooo
12 and 2
\[ y^2 -24= 2y^2 -14y \\ y^2 -14y + 24 = 0 \] that factors
its 12 and 2
now that we are over that snafu... find the slope of the tangent line using m = 6/y (from the derivative way up top) we know (2,7) is a point on the line, so we can use the point slope form: y - y0= m ( x - x0) where m is the slope, and (x0,y0) is a point on the line rearrange into standard form to match your answers up top
slope is 6/7
y-7 = 6/7 (x-2)
no, use the y values we found from \[ y^2 -14y + 24 = 0 \]
so 6/12 or 1/2
and 6/2 or 3
yes. you will now get two different equations for the 2 lines
got cha!! THANKS :)))))))))))
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