For the equation x^2 / 25+y^2 / 9=1, find the y coordinates when x=2 ; b) x=3 ; c) x=4 ; d) x=5

step 1 Let's look at part A. 3x plus 2 over x plus 5. Okay, to get the derivative, we have to write the

For the equation x^2 / 25+y^2 / 9=1, find the y coordinates...

Key Concepts

Domain and Validity in Radical Equations

When solving for y by isolating the variable and taking the square root, it is essential to ensure that the expression inside the square root is non-negative. This introduces restrictions on the x-values that can yield real y-coordinates, which is a critical consideration when working with the geometry of ellipses.

Substitution Method in Equations

Substitution is a fundamental algebraic technique where a known value is substituted into an equation to solve for an unknown variable. In the context of conic sections, substituting a specific x-value into the ellipse equation allows for solving the corresponding y coordinates, provided that the resulting expression is defined.

Conic Sections - Ellipse

An ellipse is a type of conic section defined as the set of points for which the sum of the distances to two fixed points (the foci) is constant. In its standard form, its equation is written in the form (x^2/a^2) + (y^2/b^2) = 1, which represents an ellipse centered at the origin with axes lengths determined by the values of a and b.

Standard Equation of an Ellipse

The standard form (x^2/a^2) + (y^2/b^2) = 1 clearly shows the scaling factors along the x- and y-axes. This form is useful because it immediately reveals the semi-major and semi-minor axes of the ellipse, allowing one to analyze the shape and location of the ellipse relative to the coordinate axes.

Transcript

00:01 Let's look at part a.

00:08 3x plus 2 over x plus 5.

00:16 Okay, to get the derivative, we have to write the denominator times the derivative of the numerator minus the numerator.

00:32 Don't forget to write your parentheses, times the derivative of the denominator, which is just 1 over denominator squared.

00:42 Now, if we put in a negative 3, then we've got negative 3 plus 2 is negative 2.

01:10 Negative 3 times 3 is negative 9 plus 2 is negative 7.

01:21 Negative 3 plus 1 is negative 2, negative 2, negative 2 squared is 4.

01:28 Okay, so that's gonna be, the numerator is going to evaluate to 1 and the denominator to four, that's going to give me one -fourth, which would be correct except for the fact that i wrote the question down wrong.

02:00 It's supposed to be an x plus five in the denominator.

02:03 So i'm going to change this to five.

02:23 Changes to five.

02:27 Changes this to five.

02:30 Okay, so now i think we're good.

02:32 Now i just have to erase this.

02:35 Erase this.

02:39 Okay.

02:42 Negative 3 plus 5 is 2.

02:49 And negative 3 plus 5 in the denominator is also 2 squared is 4.

02:54 So now that gives me 13 when i evaluate it.

03:00 And that's the correct answer.

03:03 B.

03:15 Make sure i write it down correctly this time.

03:18 Okay.

03:18 The derivative is going to be the denominator times the derivative of the numerator, which would be 2x squared, minus the numerator, times the derivative of the denominator, which would be 2x, over denominator squared.

03:51 If i substitute in 1, well, that's going to give me 1 squared plus 9 is 10, 2 times 2 times 1 squared is 2.

04:07 Come on right already.

04:13 Minus 1 cubed is 1 times 2 times 1 is 2.

04:21 And 1 squared plus 9 is 10.

04:26 Squared is 100.

04:38 So that gives me 20 minus 2 is 18 over 100.

04:50 Should be 9 .50ths.

05:04 And i made another mistake.

05:06 I notice that right here that's supposed to be a three.

05:12 So it's going to change that to three, change that to three, because the derivative of x cubed, 3x squared.

05:31 So that's going to be 30 minus 2, which will be 28, and then i divide by 4, and i get 7, 25ths...