The equations of two lines through the point (1, 16) tangent to the curve of
4x²-8x+ y²+2=0
are given by
y=m₁x+b₁ and y = m2x+b2
Then m1* m2 =
AAAaannndd the bot gets it wrong yet again!
Blew the very first step ...
8x - 8 + 2y(dy/dx) = 0
dy/dx = (4 - x) / y
It should have been
4x^2 - 8x + y^2 + 2 = 0
8x - 8 + 2yy' = 0
y' = (8-8x)/2y = (4-4x)/y
so at (1,16) y' = 0
well, that makes it easy. You just want the two ends of the major axis of the ellipse
(x-1)^2/(1/2) + y^2/2 = 1
That is, the points (1,±√2)
so the two tangent lines are the horizontal lines y=±√2
First, we can find the derivative of the given equation to get the slope of the tangent line at any point:
8x - 8 + 2y(dy/dx) = 0
dy/dx = (4 - x) / y
For the tangent lines to pass through (1,16), we can substitute x=1 and y=16 to get the slope of the tangent lines:
dy/dx = (4-1) / 16 = 3/16
So the equations of the tangent lines at (1,16) are:
y - 16 = (3/16)(x - 1) => y = (3/16)x + 15.25
and
y - 16 = -(16/3)(x - 1) => y = -(16/3)x + 64.33
The product of the slopes of these lines is:
m1 * m2 = (3/16) * (-16/3) = -1
Therefore, the answer is -1.