Module 5: Coordinate Geometry

Dear Class 12 Student! Coordinate Geometry is a very important branch of Mathematics. It is used in many topics of Class 12, especially in Calculus (Area under Curves, Differential Equations). A solid foundation in Straight Lines, Circles, and Conic Sections is essential for mastering advanced mathematics. Let's master the foundation!

Topic 4 Check (Part A) Question: For the ellipse $\frac{x^2}{16} + \frac{y^2}{4} = 1$, find the coordinates of the vertices and determine its orientation.

Solution: Comparing with $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$:
1. $a^2 = 16 \implies a = 4$. Crosses x-axis at (4, 0) and (-4, 0).
2. $b^2 = 4 \implies b = 2$. Crosses y-axis at (0, 2) and (0, -2).
3. Since $a \gt b$, it is a horizontal ellipse.

Topic 4 Check (Part B) Question: Given the non-standard ellipse equation $9x^2 + 16y^2 = 144$, find the lengths of the major and minor axes.

Solution: Divide the entirety by $144$ to yield standard form: $\frac{x^2}{16} + \frac{y^2}{9} = 1$.
Here $a^2 = 16 \implies a = 4$ and $b^2 = 9 \implies b = 3$.
Since $a \gt b$, it is a horizontal ellipse.
Major axis = $2a = \mathbf{8}$. Minor axis = $2b = \mathbf{6}$.

MEGA PRACTICE BANK (Advanced Conics)

Problem 5: Intersection of Two Figures Question: Find the points of intersection between the circle $x^2 + y^2 = 8$ and the straight line $y = x$.

Solution: Substitute $y = x$ into the circle equation: $x^2 + (x)^2 = 8 \implies 2x^2 = 8 \implies x^2 = 4 \implies x = \pm 2$.
Since $y = x$, when $x = 2$, $y = 2$. When $x = -2$, $y = -2$.
Intersection coordinates: $\mathbf{(2, 2)}$ and $\mathbf{(-2, -2)}$.

Problem 6: Area Bounding Preparation Question: State the exact points bounded horizontally where the parabola $y = x^2$ and the line $y = 4$ intersect.

Solution: Set the graphs equal to each other algebraically: $x^2 = 4 \implies x = \pm 2$.
The bounded geometric region lies tightly between the curve and the line spanning from $\mathbf{x = -2 \text{ to } x = 2}$. These are the boundaries you will use for Class 12 Area integrations!

Problem 7: Shifting Axes Analysis Question: What conic section does the equation $x^2 + y^2 - 2x + 4y + 5 = 0$ represent? Identify it.

Solution: Method: Complete the square for both $x$ and $y$ variables:
$(x^2 - 2x + 1) + (y^2 + 4y + 4) = -5 + 1 + 4$
$(x - 1)^2 + (y + 2)^2 = 0$
Because the radius calculation $r^2$ is exactly $0$, it forms a point circle at $(1, -2)$, rather than drawing a tangible curve area.