Respuesta :
Answer:
The tangent line equations are:
- The one that has the smaller slope [tex]\bold{y =3}[/tex]
- The one that has the larger slope [tex]\bold{y = \cfrac 14 x - \cfrac{15}{4}}[/tex]
Step-by-step explanation:
In order to find the tangent lines we need to find first the first derivative since the first derivative evaluated at a point give us the slope of the tangent line.
Working with implicit differentiation.
In order to find the derivative we can work with implicit differentiation and we get
[tex]2x+18y \cfrac{dy}{dx}= 0[/tex]
Notice that only when we are finding the derivative of an expression that has y we multiply by dy/dx due chain rule.
Solving for the first derivative we get
[tex]18y\cfrac{dy}{dx}=-2x[/tex]
[tex]\cfrac{dy}{dx}=-\cfrac{x}{9y}[/tex]
So the equation of the first derivative at the point (x,y) give us the first equation for the slope.
[tex]m=-\cfrac{x}{9y}[/tex]
Slope of a line using definition.
We can also find the slope of the line that will pass the point (x,y) trough the point (27,3) using the definition:
[tex]m = \cfrac{y_2-y_1}{x_2-x_1}[/tex]
Replacing the point we get
[tex]m=\cfrac{y-3}{x-27}[/tex]
That is another equation for the slope.
Finding the value of the slopes.
We can now set both slope equations equal to each other to find the point (x,y) where they intersect and the value of the slopes.
[tex]-\cfrac{x}{9y}= \cfrac{y-3}{x-27}[/tex]
We can work with cross multiplication to get
[tex]-x(x-27)=9y(y-3)[/tex]
And we can distribute and simplify
[tex]-x^2+27x=9y^2-27y[/tex]
Moving everything to the right side
[tex]0=x^2-27x+9y^2-27y\\\\0=x^2+9y^2-27x-27y[/tex]
At this point we can use the ellipse equation so we can replace [tex]x^2+9y^2[/tex] with 81, that will give us
[tex]0=81-27x-27y[/tex]
Then we can divide by 27 and solve for y.
[tex]0=3-x-y\\y= 3-x[/tex]
At this point we can replace on the ellipse equation.
[tex]x^2+9(3-x)^2=81[/tex]
And we can distribute and simplify.
[tex]x^2+9(9-6x+x^2)=81\\x^2+81-54x+9x^2=81\\x^2-54x+9x^2=0\\10x^2-54x=0\\[/tex]
So then we can factor and solve for x.
[tex]2x(5x-27)=0[/tex]
Setting each factor and solve for x we get
[tex]x= 0, \cfrac{27}{5}[/tex]
At x = 0, we can find the y value.
[tex]y= 3-x\\y= 3-0\\y= 3[/tex]
And the slope
[tex]m = -\cfrac{x}{9y}\\m = -\cfrac{0}{9y}\\m = 0[/tex]
So the line equation is
[tex]y-3=0(x-0)\\y=3[/tex]
For the second point x = 27/5 we have:
[tex]y= 3-\cfrac{27}{5}\\y=-\cfrac{12}{5}[/tex]
So slope is:
[tex]m = -\cfrac{\cfrac{27}{5}}{9\left(-\cfrac{12}{5}\right)}[/tex]
[tex]m = \cfrac 14[/tex]
So the line equation is
[tex]y-\left(-\cfrac{12}{5}\right)= \cfrac 14 \left(x -\cfrac{27}{5}\right)\\y = \cfrac 14 x - \cfrac{15}{4}[/tex]
Thus the equations of tangent lines are:
[tex]\bold{y =3}[/tex]
and
[tex]\bold{y = \cfrac 14 x - \cfrac{15}{4}}[/tex]
The equation of a tangent line passing through the point (27, 3) is [tex]2x + y = 57[/tex]
What is a tangent line?
The line which touches the curve at a single point is called a tangent line.
The given curve of an ellipse is [tex]\rm x^{2} + 9y^2 = 81[/tex].
And the line is passing through the point (27, 3).
Differentiate the curve with respect to x.
[tex]\begin{aligned} \dfrac{d}{dx} \rm (x^{2} + 9y^2 ) &= \dfrac{d}{dx}81\\\\2x + 18y \dfrac{dy}{dx} &= 0\\\\\dfrac{dy}{dx} &= \dfrac{-2x}{18y} = \dfrac{-x}{9y}\\\\\dfrac{dy}{dx} &= \dfrac{-2*27}{9*3} = -2\end{aligned}[/tex]
The slope of the line is -2. Then the equation of a line passing through the point (27, 3) and having a slope is -2.
[tex]\rm y\ -\ 3\ = -2(x - 27)\\\\2\ x + y = 57[/tex]
Thus, the equation of a tangent line passing through the point (27, 3) is [tex]2x + y = 57[/tex].
More about the tangent line link is given below.
https://brainly.com/question/17060643
