Find the equation of the tangents to the curve x^2+3x-2y^2=4 at the points where the curve crosses the x- axis
To find the equation of the tangents to the curve at the points where it crosses the x-axis, we need to follow these steps:
Step 1: Find the x-intercepts of the curve.
The x-intercepts are the points where the curve crosses the x-axis. To find these points, set y = 0 in the equation x^2 + 3x - 2y^2 = 4 and solve for x:
x^2 + 3x - 2(0)^2 = 4
x^2 + 3x = 4
x^2 + 3x - 4 = 0
Now we have a quadratic equation. We can solve it by factoring or using the quadratic formula. In this case, let's use factoring:
(x + 4)(x - 1) = 0
Setting each factor equal to zero gives us:
x + 4 = 0 or x - 1 = 0
So the x-intercepts are x = -4 and x = 1.
Step 2: Find the corresponding y-coordinates for the x-intercepts.
To find the corresponding y-coordinates for the x-intercepts, substitute the x-values into the original equation of the curve: x^2 + 3x - 2y^2 = 4.
For x = -4:
(-4)^2 + 3(-4) - 2y^2 = 4
16 - 12 - 2y^2 = 4
-2y^2 = 0
y^2 = 0
y = 0
So the corresponding y-coordinate for x = -4 is y = 0.
For x = 1:
(1)^2 + 3(1) - 2y^2 = 4
1 + 3 - 2y^2 = 4
-2y^2 = 0
y^2 = 0
y = 0
So the corresponding y-coordinate for x = 1 is also y = 0.
Step 3: Find the slopes of the tangents.
The slope of a tangent line to the curve at a given point is equal to the derivative of the curve at that point.
To find the derivative, implicitly differentiate the equation x^2 + 3x - 2y^2 = 4 with respect to x:
2x + 3 - 4y * (2y') = 0
Simplifying further:
2x + 3 - 8yy' = 0
8yy' = 2x + 3
y' = (2x + 3) / (8y)
Step 4: Substitute the x and y values of each x-intercept and find the slopes of the tangents.
For x = -4 and y = 0, we have:
y' = (2(-4) + 3) / (8(0))
y' = -5/0
The slope is undefined. This means that the tangent is a vertical line at x = -4.
For x = 1 and y = 0, we have:
y' = (2(1) + 3) / (8(0))
y' = 5/0
Again, the slope is undefined. This means that the tangent is a vertical line at x = 1.
Step 5: Write the equations of the tangents.
Since the slopes are undefined, the equations of the tangents are simply x = -4 and x = 1.
Therefore, the equations of the tangents to the curve x^2 + 3x - 2y^2 = 4 at the points where the curve crosses the x-axis are x = -4 and x = 1.