Question

Help with calculus problem about implicit differentiation for tangent lines at an ellipse given a point.

Answer 1

`\textsf{Smaller slope:}\quad y=3`

`\textsf{Larger slope:}\quad y=(2)/(7)x-(27)/(7)`

Explanation:

To find the equations of the tangent lines to the ellipse x² + 8y² = 72 that pass through the point (24, 3), we can use the implicit differentiation method for finding the slopes of the tangents.

To differentiate an equation that contains a mixture of x and y terms, begin by placing d/dx in front of each term of the equation:

`\frac{\text{d}}{\text{d}x}(x^2) + \frac{\text{d}}{\text{d}x}(8y^2) = \frac{\text{d}}{\text{d}x}(72)`

Differentiate the terms in x only and the constant:

`2x + \frac{\text{d}}{\text{d}x}(8y^2) = 0`

Use the chain rule to differentiate terms in y. In practice, this means differentiate with respect to y, and place dy/dx at the end:

`2x + 16y\frac{\text{d}y}{\text{d}x} = 0`

Rearrange to make dy/dx the subject:

`\frac{\text{d}y}{\text{d}x} = (-2x)/(16y)`

`\frac{\text{d}y}{\text{d}x} = -(x)/(8y)`

Therefore, the slope of the tangent line at a point (x, y) on the ellipse is given by:

`\frac{\text{d}y}{\text{d}x} = -(x)/(8y)`

The slope of a line that passes through points (x, y) and (24, 3) is:

`m=(3-y)/(24-x)`

Equate the two slopes:

`(3-y)/(24-x)=-(x)/(8y)`

`8y(3-y)=-x(24-x)`

`24y-8y^2=-24x+x^2`

`24y=x^2-24x+8y^2`

Rearrange the original equation x² + 8y² = 72 to isolate 8y², then substitute it into the equation:

`24y=x^2-24x+(72-x^2)`

`24y=-24x+72`

`y=3-x`

Substitute y = 3 - x into the original equation of the ellipse, then solve for x to find the x-coordinates of the points of tangency of the ellipse:

`x^2 + 8(3 - x)^2 = 72`

`x^2 + 8(9-6x+x^2) = 72`

`x^2 + 72-48x+8x^2 = 72`

`9x^2 -48x=0`

`3x(3x -16)=0`

Therefore, the x-coordinates of the points of tangency are:

`3x=0 \implies x=0`

`3x-16=0 \implies x=(16)/(3)`

To find the y-coordinates, substitute the x-coordinates into the equation y = 3 - x:

`\begin{aligned}x=0 \implies y&=3-0\\y&=3\end{aligned}`

`\begin{aligned}x=(16)/(3) \implies y&=3-(16)/(3)\\y&=-(7)/(3)\end{aligned}`

Therefore, the points of tangency are:

`(0,3)\;\;\textsf{and}\;\;\left((16)/(3),-(7)/(3)\right)`

Substitute the points of tangency into the dy/dx slope expression to find the slopes of the two tangent lines:

`(0,3)\implies m_1=(3-3)/(24-0)=(0)/(24)=0`

`\left((16)/(3),-(7)/(3)\right)\implies m_2=(3-\left(-(7)/(3)\right))/(24-(16)/(3))=((16)/(3))/((56)/(3))=(2)/(7)`

Finally, substitute the found slopes and point (24, 3) into the point-slope form of a linear equation to create the equations of the tangent lines:

`\begin{aligned}\textsf{Equation 1:}\quad y-3&= 0\left(x-24)\\\\y&=3\end{aligned}`

`\begin{aligned}\textsf{Equation 2:}\quad y-3&= (2)/(7)\left(x-24)\\\\y&=(2)/(7)x-(27)/(7)\end{aligned}`