Problem:
Find the point of intersection of tangents at P and Q for the given equations: 3x + 4y = 12 and x^2/25 + y^2/16 = 1.

Solution:

Step 1: Find the coordinates of points P and Q by solving the given equations simultaneously.

First, let's solve the equation 3x + 4y = 12 for y:
4y = 12 - 3x
y = (12 - 3x)/4

Step 2: Substitute the value of y into the equation of the ellipse and solve for x:

(x^2/25) + ((12 - 3x)/4)^2 = 1

Multiplying both sides by 100 to eliminate the denominators:
4x^2 + 25(12 - 3x)^2 = 100

Expanding and simplifying:
4x^2 + 25(144 - 72x + 9x^2) = 100
4x^2 + 3600 - 1800x + 225x^2 = 100
229x^2 - 1800x + 3500 = 0

Step 3: Solve the quadratic equation:

Using the quadratic formula, x = (-b ± √(b^2 - 4ac))/2a, where a = 229, b = -1800, and c = 3500.

x = (-(-1800) ± √((-1800)^2 - 4(229)(3500)))/(2(229))
x = (1800 ± √(3240000 - 3214000))/(458)
x = (1800 ± √26000)/458

Calculating the square root:
x = (1800 ± 160.62)/458

Simplifying:
x = (1800 + 160.62)/458 or x = (1800 - 160.62)/458
x = 3.96 or x = 3.21

Step 4: Substitute the values of x back into the equation 3x + 4y = 12 to find the corresponding y-values:

For x = 3.96:
3(3.96) + 4y = 12
11.88 + 4y = 12
4y = 12 - 11.88
4y = 0.12
y = 0.03

For x = 3.21:
3(3.21) + 4y = 12
9.63 + 4y = 12
4y = 12 - 9.63
4y = 2.37
y = 0.59

Step 5: Calculate the slopes of the tangents at points P and Q:

The slope of the tangent at P is the derivative of the ellipse equation at point P. Differentiating x^2/25 + y^2/16 = 1 with respect to x:
(2x/25) + (2y/16) * (dy/dx) = 0
(2x/25) + (2y/16) * (dy/dx

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