Find the equation of tangent to the ellipse 4x^2+5y^2=20 which are perpendicular to the line 3x+2y-5=0
take the derivative of the ellipse equation using implicit differentiation.
The slope of 3x+2y - 5 = 0 is -3/2
so the slope of the perpendicular will be +2/3
set your derivative of the ellipse equal to 2/3 .
You will get a linear equation in x and y
Solve that with the original ellipse equation.
(I would use substituion)
To find the equation of the tangent to the ellipse that is perpendicular to the given line, we need to follow these steps:
Step 1: Find the center and axes of the ellipse.
Step 2: Determine the slope of the given line.
Step 3: Find the normal vector to the line.
Step 4: Find the gradient of the ellipse at the point of tangency.
Step 5: Find the equation of the tangent line using the point of tangency and the gradient.
Let's break down each step in detail:
Step 1: Find the center and axes of the ellipse.
The given equation of the ellipse is 4x^2 + 5y^2 = 20. By comparing this equation with the standard form of an ellipse equation x^2/a^2 + y^2/b^2 = 1, we can conclude that the center is at the origin (0, 0) and the semi-major axis (a) is √5 and the semi-minor axis (b) is 2.
Step 2: Determine the slope of the given line.
The given line has the equation 3x + 2y - 5 = 0. We can rewrite this equation in slope-intercept form (y = mx + c) by isolating y. The slope (m) of the line is then -3/2.
Step 3: Find the normal vector to the line.
The normal vector to a line with slope m is the negative reciprocal of the slope, so the normal vector to the given line is 2/3.
Step 4: Find the gradient of the ellipse at the point of tangency.
The general equation of an ellipse is given by x^2/a^2 + y^2/b^2 = 1. The derivative of this equation gives the slope at any point on the ellipse.
Differentiating the equation 4x^2 + 5y^2 = 20 with respect to x, we get:
8x + 10yy' = 0
Simplifying and solving for y', we have:
y' = -8x / (10y)
The slope of the ellipse at any point (x, y) is given by y' = -8x / (10y).
Step 5: Find the equation of the tangent line using the point of tangency and the gradient.
The slope of the tangent line must be perpendicular to the slope of the given line, which means the product of their slopes must be -1.
Let the point of tangency on the ellipse be (h, k). We know that the slope of the given line is -3/2, and the slope of the tangent line at (h, k) is given by y' = -8h / (10k).
Therefore, we have:
(-3/2) * (-8h / 10k) = -1
Simplifying, we get:
12h = 15k
This equation determines the relationship between h and k.
Finally, we can substitute the value of (h, k) into the equation of the ellipse equation 4x^2 + 5y^2 = 20 to find the specific values.