AI Math Solver
Resources
Questions
Pricing
Login
Register
Home
>
Questions
>
Find the y-coordinate of a point on the unit circle given x = √35/6
Mathematics
Grade 10 of Junior High School
Question Content
The point $P = \\left( \\frac{\\sqrt{35}}{6} , y \\right)$ lies on the unit circle shown below. What is the value of $y$ in simplest form?
Correct Answer
$y = \\frac{1}{6}$
Detailed Solution Steps
1
Step 1: Recall the equation of a unit circle, which is $x^2 + y^2 = 1$, where $(x,y)$ is any point on the circle.
2
Step 2: Substitute $x = \\frac{\\sqrt{35}}{6}$ into the unit circle equation: $\\left(\\frac{\\sqrt{35}}{6}\\right)^2 + y^2 = 1$.
3
Step 3: Calculate $\\left(\\frac{\\sqrt{35}}{6}\\right)^2$: this equals $\\frac{35}{36}$ because squaring a square root cancels the root, and $6^2=36$.
4
Step 4: Rearrange the equation to solve for $y^2$: $y^2 = 1 - \\frac{35}{36}$.
5
Step 5: Compute $1 - \\frac{35}{36} = \\frac{36}{36} - \\frac{35}{36} = \\frac{1}{36}$.
6
Step 6: Take the positive square root of $\\frac{1}{36}$ (since the point $P$ is in the first quadrant as shown in the diagram, $y$ is positive) to get $y = \\frac{1}{6}$.
Knowledge Points Involved
1
Equation of the Unit Circle
The unit circle is a circle centered at the origin $(0,0)$ with a radius of 1. Its standard equation is $x^2 + y^2 = 1$, which comes from the distance formula (the distance from any point $(x,y)$ on the circle to the origin is equal to the radius 1). It is used to relate coordinates of points on the circle to trigonometric functions and solve for unknown coordinates.
2
Squaring Square Roots
For a non-negative real number $a$, $(\\sqrt{a})^2 = a$. This rule simplifies calculations involving squared radical terms, as the square root and square operations cancel each other out. It is commonly used in algebra and coordinate geometry problems involving circles or radicals.
3
Solving for a Variable in a Quadratic Equation
When solving for a variable in an equation like $x^2 + y^2 = 1$, we isolate the term with the unknown variable by rearranging the equation, then use square roots to solve for the variable. We must consider the sign of the solution based on the context (e.g., quadrant of the point in coordinate geometry).
Loading solution...
