Q.Find the equation of the parabola with vertex at $(0, 0)$ and focus at $(0, 2)$.

Concept & Intuition

Comparing $x^2=\lambda y$ with the standard form $x^2=4ay$ gives $4a=\lambda$, so $a=\lambda/4$. The focus is at $(0,a)$ and the latus rectum endpoints are $(\pm 2a, a)$. The vertex is the origin.

Step-by-step solution

1. Latus rectum endpoints in terms of $\lambda$.

   $2a = \lambda/2$ and $a=\lambda/4$, so the endpoints are $\left(-\dfrac{\lambda}{2}, \dfrac{\lambda}{4}\right)$ and $\left(\dfrac{\lambda}{2}, \dfrac{\lambda}{4}\right)$.

2. Set up the triangle.

   Vertex $(0,0)$, base $=\lambda$ (distance between the endpoints), height $=\lambda/4$ (vertical distance from the vertex to the latus-rectum line). …

Concept & Intuition

The latus rectum of a parabola is the chord through the focus perpendicular to the axis. For $y^2 = 4ax$, its endpoints are $(a, 2a)$ and $(a, -2a)$. The "negative end" means the one with negative $x$? No — here "negative end" refers to the end with a negative $y$-coordinate (since the vertex is at the origin and the latus rectum lies to the right). But wait: the problem says "passing through its vertex and the negative end of the latus rectum." The vertex is $(0,0)$. The latus rectum endpoints are $(a, 2a)$ and $(a, -2a)$. The "negative end" is $(a, -2a)$. However, the chord goes from $(0,0)$ to $(a, -2a)$. That gives slope $\frac{-2a}{a} = -2$, so equation $y = -2x$ or $2x + y = 0$. But we must first find $a$ from the given length.

Step-by-step

1. Find $a$ from the latus rectum length For $y^2 = 4ax$, the length of the latus rectum is $4a$. Given that this length is 6 units:

2. Identify the endpoints of the latus rectum

   With $a = 1.5$, the endpoints are $(1.5, 3)$ and $(1.5, -3)$. The "negative end" means the one with the negative $y$-coordinate: $(1.5, -3)$.

3. Find the chord through the vertex and this point …

[!TLDR]

Focus (0,-2) with directrix y=2 gives a downward parabola $x^2=-8y$.

Concept

For a vertical parabola with vertex at the origin, $x^2 = -4ay$ opens downward, with focus at $(0,-a)$ and directrix $y=a$ (CBSE/NCERT Class 11, Conic Sections).

Solution …

The key here is to connect the given point to the parameter $a$ of the parabola. The standard form $x^2 = 4ay$ tells us two things: the parabola opens upward, and the length of the latus rectum (the chord through the focus perpendicular to the axis) is exactly $4a$. So the problem reduces to finding $a$ from the condition that $(2,1)$ lies on the curve.

1. Substitute the point into the equation. Since $(2,1)$ satisfies $x^2 = 4ay$, we plug $x = 2$ and $y = 1$:

2. Solve for $a$.

   Dividing both sides by $4$ gives $a = 1$.

3. Recall the latus rectum length.

   For any parabola $x^2 = 4ay$, the latus rectum is the line segment through the focus $(0, a)$ parallel to the $x$-axis, with endpoints at $x = \pm 2a$. Its length is $4a$.

   With $a = 1$, the length is $4 \times 1 = 4$. …