Find the equation of the tangent line to the ellipse x 2 + 4y 2 = 25 when x = 3 and y < 0.
Answer
3x - 8y = 7
3x - 8y = 25
3x + 8y = 7
3x + 8y = 25
8 y dy + 2 x dx = 0
so
dy/dx = m = -x/4y
at x = 3, y = -2
dy/dx = -3/(4y) = 3/8 = m
so
y = (3/8) x + b
-2 = (3/8)3 + b
b = -25/8
so
y = (3/8) x -25/8
8 y - 3 x = -25
3 x - 8 y = 25 (b)
To find the equation of the tangent line to the ellipse when x = 3 and y < 0, we can use the concept of implicit differentiation.
First, let's differentiate the equation of the ellipse with respect to x to find the derivative of y with respect to x:
d/dx (x^2 + 4y^2) = d/dx (25)
2x + 8y(dy/dx) = 0
Now, we can substitute x = 3 into the equation above:
2(3) + 8y(dy/dx) = 0
6 + 8y(dy/dx) = 0
Next, we are given that y < 0. Since we are looking for the tangent line when x = 3 and y < 0, we substitute this information into the equation:
6 + 8(y)(dy/dx)|_(y<0) = 0
Simplifying further, we have:
8y(dy/dx)|_(y<0) = -6
Dividing both sides by 8y, we get:
(dy/dx)|_(y<0) = -6 / 8y
Now, we can determine the slope of the tangent line at the point (3, y) when y < 0 by substituting y into the equation above.
Next, we can evaluate dy/dx at the given point (3, y) and find the slope of the tangent line.
Finally, we can use the point-slope form of a line (y - y1 = m(x - x1)) using the slope we found and the point (3, y) to find the equation of the tangent line.