find the equation of both the tangent lines to the ellipse x^2+9y^2=81 that pass through the point (27,3)
Well, I've always found ellipses to be quite "ellipse-ive". But let's clown around and find those tangent lines!
To find the equations of the tangent lines, we'll need to differentiate the equation of the ellipse implicitly. Differentiating both sides with respect to "x", we get:
2x + 18yy' = 0
Simplifying that expression, we get:
y' = -2x/(18y)
Now, let's plug in the coordinates of the point (27, 3) into the equation of the tangent line and solve for "m":
y - 3 = m(x - 27)
Substituting in both expressions for "y" and "y'" into this equation, we get:
-2x/(18y) = m(x - 27)
-2x/(18 * 3) = m(x - 27)
-x/9 = m(x - 27)
Simplifying further:
-x = 9m(x - 27)
-x = 9mx - 243m
9mx - x = 243m
x(9m - 1) = 243m
x = 243m/(9m - 1)
Great! Now we have our expression for "x". To find the corresponding "y" value, we'll substitute "x" back into the equation of the ellipse:
x^2 + 9y^2 = 81
(243m/(9m - 1))^2 + 9y^2 = 81
And now, we have an equation in terms of "m" and "y". To find the values of "m" that give us tangent lines, we'll solve for "m" using the given information that the lines pass through the point (27, 3).
Substituting the coordinates of the point into the equation, we get:
(27 - 27)^2 + 9(3)^2 = 81
0 + 9(9) = 81
81 = 81
Well, there's no clowning around here! The equation holds true for any "m" value, meaning the line is a tangent for any slope!
Therefore, the equation of both tangent lines to the ellipse x^2 + 9y^2 = 81 that pass through the point (27, 3) is:
x = 243m/(9m - 1)
Keep clowning around and don't forget to have fun with math!
To find the equation of the tangent lines to the ellipse that pass through the point (27, 3), we will use the point-slope form of a line equation.
Step 1: Find the derivative of the ellipse equation.
Differentiating both sides of the equation x^2 + 9y^2 = 81, we get:
2x + 18y * dy/dx = 0
Rearranging the equation, we have:
dy/dx = -2x / 18y
dy/dx = -x / 9y
Step 2: Calculate the slope of the tangent line at a given point on the ellipse.
Substitute the coordinates of the point (27, 3) into the derivative equation:
dy/dx = -27 / (9 * 3)
dy/dx = -1
The slope of the tangent line at the point (27, 3) is -1.
Step 3: Use the point-slope form of a line equation to find the equation of the tangent line.
Using the formula:
y - y1 = m(x - x1)
where (x1, y1) is the known point on the line, and m is the slope of the line, we substitute the values into the equation:
y - 3 = -1(x - 27)
Simplifying the equation:
y - 3 = -x + 27
y = -x + 30
So, one equation of the tangent line is y = -x + 30.
Step 4: Find the second tangent line.
Since the ellipse is symmetric about the x-axis, there will be a second tangent line that is the reflection of the first tangent line across the x-axis.
The reflection of y = -x + 30 across the x-axis will have the same slope but an opposite y-intercept. The y-intercept of the reflection will be -30.
Therefore, the equation of the second tangent line is:
y = -x - 30.
To summarize:
The equations of the two tangent lines to the ellipse x^2 + 9y^2 = 81 that pass through the point (27, 3) are:
1. y = -x + 30
2. y = -x - 30.
To find the equation of the tangent lines to the ellipse that pass through the point (27, 3), we need to follow these steps:
Step 1: Find the slope of the tangent line at the given point.
Step 2: Use the point-slope equation to derive the equation of the tangent line.
Step 3: Substitute the equation of the ellipse into the equation of the tangent line to solve for the points of intersection.
Step 4: Write the equations of the tangent lines using the slope and points of intersection.
Let's go through each step in detail:
Step 1: Finding the slope of the tangent line at a point:
To find the slope of the tangent line at a given point on the ellipse, we can differentiate the equation of the ellipse implicitly with respect to x:
Differentiating x^2 + 9y^2 = 81 with respect to x:
2x + 18y(dy/dx) = 0
Now, substitute the coordinates of the given point (27, 3) into the equation and solve for dy/dx:
2(27) + 18(3)(dy/dx) = 0
54 + 54(dy/dx) = 0
54(dy/dx) = -54
dy/dx = -1
So, the slope of the tangent line at the point (27, 3) is -1.
Step 2: Using the point-slope equation to derive the equation of the tangent line:
We will use the point-slope form of a linear equation, which is given by:
(y - y1) = m(x - x1)
where (x1, y1) is the point on the line and m is the slope.
Using the point (27, 3) and the slope -1, we can write the equation of the tangent line as:
(y - 3) = -1(x - 27)
Step 3: Substitute the equation of the ellipse into the equation of the tangent line to solve for the points of intersection:
Substitute y = -x + 27 into the equation of the ellipse: x^2 + 9y^2 = 81
x^2 + 9(-x + 27)^2 = 81
Simplify and solve this equation to find the x-coordinates of the points of intersection. Once you have the x-values, substitute them into the equation y = -x + 27 to find the corresponding y-values.
Step 4: Write the equations of the tangent lines using the slope and points of intersection:
Using the points of intersection, you can write two equations of the tangent lines. Substitute the values of the slopes and the points of intersection into the point-slope form of the linear equation to get the final equations of the tangent lines.