(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 $\frac{x^{2}}{2}+\frac{y^{2}}{8}=1$ at $(1,2)$ . (b) Show that the equation of the tangent line to the ellipse $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$ at $\left(x_{0}, y_{0}\right)$ is $\frac{x_{0} x}{a^{2}}+\frac{y_{0} y}{b^{2}}=1$

(a) Use implicit differentiation to find an equation of the tangent line to the ellipse $\frac{x^{2}}{2}+\frac{y^{2}}{8}=1$ at $(1,2)$ .
(b) Show that the equation of the tangent line to the ellipse $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$ at $\left(x_{0}, y_{0}\right)$ is $\frac{x_{0} x}{a^{2}}+\frac{y_{0} y}{b^{2}}=1$

Key Concepts

General Tangent Line Equation for an Ellipse

For an ellipse in the standard form, the equation of the tangent line at a point (x?, y?) on the ellipse can be derived by utilizing the properties of the ellipse and its symmetry. The resulting formula, (x?x)/a² + (y?y)/b² = 1, provides a quick method to find the tangent line without resorting to implicit differentiation each time.

Standard Form of an Ellipse

An ellipse can be expressed in its standard form as x²/a² + y²/b² = 1. This form highlights its geometric properties, such as the lengths of its semi-major and semi-minor axes, and serves as the starting point for many analytical techniques, including finding tangent lines.

Implicit Differentiation

This is a method used to differentiate equations in which y is given implicitly as a function of x. Instead of solving for y explicitly, one differentiates both sides of the equation with respect to x, treating y as an implicit function of x, and then solves for dy/dx.

Tangent Line to a Curve

The tangent line to a curve at a given point is the line that just touches the curve at that point and has the same slope as the curve there. Its equation is usually obtained using the point-slope form by plugging in the point of tangency and the derivative of the function at that point.

Transcript

00:01 This is problem number 41 in which we need to use implicit differentiation to find in part a, to find the equation of the tangent line to the ellipse given ellipse is x squared by 2 plus y square by 8 equal to 1 and point is 1 .2.

00:19 So if this is let us say x not, this is y0, so equation of ellipse will be y minus y not equal to d y by d x at x0 comma y0 x minus x not where this quantity is the equation or is the value of the slope of the tangent at point x0 comma y not so to get the slope of tangent we need to find first divide by d x so differentiate the given equation using implicit differentiation using our okay implicit differentiation so we have to differentiate both sides by with respect to x equal to d by d x 1 okay so this can be written as d x square by you should write 1 by 2 here d x plus 1 by 8 t of y square by d x equal to 0 because 1 is constant and it has differentiation equal to 0 so this will be 1 by 2 into 2x plus 1 by 1 by 8 into 2 y and d y by and d y by d x equal to 0 now let us multiply both sides by okay before multiplying we should just cancel this out so it will be x plus y by 2 okay here it should be y by 4 not y by 2 y by 4 okay y by d x equal to 0 now let us subtract x from both the sides we will be getting by 4, d .y by d x equal to minus x and let us multiply both sides by four by y.

02:23 We will be just having d .y by dx equal to minus 4x by y.

02:31 This is the general term for dy by dx.

02:36 Now we have to find d .y by dx at x not y not, that is 1 .2 minus 4 into 1.

02:46 Divide by y is 2 so this will be minus 2 so equation of tangent will be y minus y not equal to minus 2 x minus x not so y minus y not will be minus 2x plus 2 now let us add 2 to both the sides so y will be equal to minus 2x plus okay so this is the equation of tangent required equation of tangent in part a now comes part b we have to show equation of tangent line to the ellipse x square by a square plus y square by b square equal to one at x0 y0 so first of all we need to find d by d x that is slope of the tangent so let us differentiate this equation equation of ellipse with respect to x to both the sides d by d x and 0 or not 0 it will be 1 okay so this is 1 by a square since this is constant can be taken out and differentiation of x square with respect to x will be 2x plus similarly 1 by b squared 2y and according to chain rule d y by d x equal to 0 because 1 will have differentiation equal to 0 so this is 2xy a square plus 2y by b square, d .y by d x equal to 0...

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