SOLUTION: The equation of the tangent line to the curve x^2 + y^2 =169 at the point (5,-12) is (A) 5y-12x= -120 (B) 5x-12y= 119 (C) 5x-12y= 169 (D) 12x+5y= 0 (E) 12x+5y= 169

Algebra ->  Functions -> SOLUTION: The equation of the tangent line to the curve x^2 + y^2 =169 at the point (5,-12) is (A) 5y-12x= -120 (B) 5x-12y= 119 (C) 5x-12y= 169 (D) 12x+5y= 0 (E) 12x+5y= 169      Log On


   

Question 243003: The equation of the tangent line to the curve x^2 + y^2 =169 at the point (5,-12) is
(A) 5y-12x= -120
(B) 5x-12y= 119
(C) 5x-12y= 169
(D) 12x+5y= 0
(E) 12x+5y= 169
Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!
First derive both sides of with respect to 'x' to get







Now solve for y':










So the slope is simply the negative quotient of the two coordinates. For the point (5,-12), the slope at that point is m=-5%2F%28-12%29=5%2F12 (since x=5 and y=-12)


-----------------------------------------------------


Now recall that the point slope formula is

y-y%5B1%5D=m%28x-x%5B1%5D%29 where 'm' is the slope and is the point in which the line goes through.


y--12=%285%2F12%29%28x-5%29 Plug in m=5%2F12, x%5B1%5D=5, and y%5B1%5D=-12


y%2B12=%285%2F12%29%28x-5%29 Rewrite y--12 as y%2B12


y%2B12=%285%2F12%29x%2B%285%2F12%29%28-5%29 Distribute 5%2F12


y%2B12=%285%2F12%29x-25%2F12 Multiply 5%2F12 and -5 to get -25%2F12


y=%285%2F12%29x-25%2F12-12 Subtract 12 from both sides to isolate y


y=%285%2F12%29x-169%2F12 Combine like terms -25%2F12 and -12 to get -169%2F12


12y=5x-169 Multiply EVERY term by the LCD 12 to clear out the fractions.


-5x%2B12y=-169 Subtract 5x from both sides.


5x-12y=169 Multiply EVERY term by -1 to make the 'x' coefficient positive.

So the equation of the tangent line is 5x-12y=169