(a) Use implicit differentiation to find an equation of the...

(a) Use implicit differentiation to find an equation of the tangent line to the ellipse x^2/2 + y^2/8 = 1 at (1,2). (b) Show that the equation of the tangent line to the ellipse x^2/a^2 + y^2/b^2 = 1 at (x_0, y_0) is x_0x/a^2 + y_0y/b^2 = 1.

Transcript

00:01 Alright, so in this problem, we are going to start off with part a.

00:03 So for part a, we are given the following equation.

00:07 We're given x squared over 2 plus y squared over 8, and this is equal to 1, and we're also given a point, so at 1 .2.

00:18 First, we are going to find, though, derivative of each of the components with respect to x.

00:23 So that's going to be 2x over 2 plus 2y over 8 times.

00:30 Dy over the x and this is going to equal to zero that means that one -fourth times y times the y over the x is going to equal to negative x which means that the y over the x when we solve for this is going to equal to negative 4x divided by y next we're going to find the slope m is going to be the value of the y at the point 1 comma 2 this is going to be the slope of the tangent line so in our case that's just going to equal to negative 4 divided by 2 which is equal to negative 2 so tangent line tangent line is going to be y minus 2 equals to the slope which is negative 2 times x minus 1 distributing negative 2 on the right hand side we get y minus 2 equals to negative 2x plus 2 and thus adding 2 on both sides of the equation gives us y -cons 2 to negative 2x plus 4 next we are going to move on to part b so proceeding on to part b here we are given the following equation we're given x squared over a squared plus y squared over b squared equals to 1 and we're also given a point at the point in general x o y o so it's a general point so we're going to do the same thing find the partial the derivatives of each of the components which will be 2x divided by a squared plus 2y divided by b squared times the y over d x and this will be equal to 0 next we want to solve for the y over d x so first 2y divided by b squared times d y over the x is going to equal to negative 2x divided by a squared and thus we get that d y over the x should be equal to negative b squared over a squared times x over y next the slope m is going to be the y over d y over the x at the point x o y o just like before so in our case that's going to be negative b squared over a squared times x o over y o so the tangent line in this case tangent line is going to be y minus y o the slope here is negative b squared over a squared times x o divided by y o times x minus x o so next we're going to distribute out a squared times y o so we are going to get the following y o times a squared times y o squared minus y o squared times a squared is going to equal to negative and we're going to distribute out what remains out...

Question

(a) Use implicit differentiation to find an equation of the tangent line to the ellipse x^2/2 + y^2/8 = 1 at (1,2). (b) Show that the equation of the tangent line to the ellipse x^2/a^2 + y^2/b^2 = 1 at (x_0, y_0) is x_0x/a^2 + y_0y/b^2 = 1.

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