the line y=x-c touches the ellipse 9x^2 + 16y^2 = 144. Find the value of c and the coordinate of the point of contact.
a parallelogram is defined by two sets of tangents to the curve x^2 - 4y^2 = 9. The gradients of the tangents are 1 and -1. Find the area of the parallelogram.
Please submit one question per post. You have posted two completely different ones here.
Substitute x-c for y in the ellipese equation. Then solve the resulting equation for x, with c as a remining unknown.
Pick the vaue of c that results in one value for x. That will correspond to the tangent-line situation.
To find the value of c and the coordinate of the point of contact for the line y=x-c and the ellipse 9x^2 + 16y^2 = 144, we can use the fact that the line is a tangent to the ellipse.
Step 1: Start by substituting y = x - c into the equation of the ellipse:
9x^2 + 16(x - c)^2 = 144
Step 2: Expand and simplify the equation:
9x^2 + 16(x^2 - 2cx + c^2) = 144
9x^2 + 16x^2 - 32cx + 16c^2 = 144
25x^2 - 32cx + 16c^2 = 144
Step 3: Since the line is a tangent, the discriminant of the quadratic equation must be zero:
Δ = b^2 - 4ac = (-32c)^2 - 4(25)(16c^2) = 0
Step 4: Solve the equation:
1024c^2 - 4(25)(16c^2) = 0
1024c^2 - 1600c^2 = 0
-576c^2 = 0
Step 5: Solve for c:
c = 0
Step 6: Substitute c = 0 back into the equation of the ellipse:
25x^2 = 144
x^2 = 144/25
x = ±12/5
Step 7: Substitute the values of x into the equation y = x - c:
When x = 12/5:
y = (12/5) - 0 = 12/5
When x = -12/5:
y = (-12/5) - 0 = -12/5
So, the value of c is 0, and the coordinates of the point of contact are (12/5, 12/5) and (-12/5, -12/5).
Now, let's move on to the second question about the parallelogram defined by two sets of tangents to the curve x^2 - 4y^2 = 9.
Step 1: Start by finding the equations of the tangents with gradients of 1 and -1.
For the gradient 1:
Differentiating the curve, we get 2x - 8yy' = 0,
where y' represents the derivative of y with respect to x.
Simplifying, we get y' = x/4y.
Since the gradient of the tangent is given, we can substitute in 1:
1 = x/4y
For the gradient -1:
Differentiating the curve, we get 2x - 8yy' = 0,
where y' represents the derivative of y with respect to x.
Simplifying, we get y' = x/4y.
Since the gradient of the tangent is given, we can substitute in -1:
-1 = x/4y
Step 2: Solving the system of equations we obtained:
From equation 1: 1 = x/4y
From equation 2: -1 = x/4y
Multiplying equation 2 by -1, we get: 1 = -x/4y
Adding this to equation 1, we have:
1 + 1 = x/4y + (-x/4y)
2 = 0
Step 3: The system of equations is inconsistent, which means there are no tangents to the curve with gradients 1 and -1.
Since there are no tangents, the parallelogram defined by these tangents does not exist. Therefore, its area is 0.