Question

find the tangents of the curve 4x^2 + 9y^2 + 24x - 18y -180 = 0 from (-6, 12). the answers are 8x + 9y - 60 = 0 and 16x - 7y + 180. i already have the process but i didnt derived to the answer. please help.

Answer 1

@phi

Answer 2

another messy algebra problem with this one.

you should get
\[ \frac{dy}{dx}= \frac{-4(x+3)}{9(y-1) } \]
and the "slope equation"
\[ \frac{-4(x+3)}{9(y-1) } = \frac{(y-12)}{(x+6)}\]

after cross multiplying
\[ -4(x+3)(x+6) = 9(y-1)(y-12) \]

you should be able to manipulate that equation into the form
\[ 4x^2 +9y^2= -36x +117y -180 \]

to simplify that we use the original equation
\[ 4x^2 + 9y^2 + 24x - 18y -180 = 0 \]
and re-arrange to get
\[4x^2 + 9y^2 = - 24x + 18y +180\]

Answer 3

now substitute into the "cross multiply" equation
\[4x^2 +9y^2= -36x +117y -180 \\ \text{ to get } \\
−24x+18y+180= -36x +117y -180 \]

solve for x or y... I solved for y to get
\[ y = \frac{4x+120}{33} \]

Answer 4

Use that value of y in the "cross multiply" equation
\[ -4(x+3)(x+6) = 9(y-1)(y-12) \\ \text{ to get }\\
-4(x+3)(x+6) = 9\left(\frac{4x+120}{33}-1\right)\left(\frac{4x+120}{33}-12\right)
\]

Answer 5

let's just concentrate on the right side
\[ 9\left(\frac{4x+120}{33}-1\right)\left(\frac{4x+120}{33}-12\right) \\
\cancel{9}3\left(\frac{4x+120-33}{\cancel{33}11}\right)\left(\frac{4x+120-396}{33}\right) \\
\cancel{3}\left(\frac{4x+87}{11}\right)\left(\frac{4x-276}{\cancel{33}11}\right) \\
\frac{4}{121}(4x+87)(x-69)


\]

Answer 6

the entire equation, using that new right hand side is
\[ -4(x+3)(x+6) = \frac{4}{121}(4x+87)(x-69) \]
divide by 4 , and multiply by 121 to get
\[ -121(x+3)(x+6)= (4x+87)(x-69) \]

Answer 7

if you work on that, you should eventually get
\[ 5x^2 +36x-153=0 \]
one of its roots is x=3 (so (x-3) divides evenly into it)

once you have the two values for x (the 2 roots) you can find y
and then the slope
and finally the equation of the tangent line through point (-6,12)

Answer 8

gaaah i am so stupid. i messed up in the cross multiplying stage.. thank you so much @phi :) thank you thank you

Answer 9

These problems require a LOT of careful work.
btw, here is a graph of the problem

Answer 10

last problem..

Answer 11

please? very last problem about this topic

Answer 12

thanks thanks :)