Question

The vertex of a parabola is the point $(a,b)$ and latus rectum is of length $1$. If the axis of the parabola is along the positive of y-axis, then its equation is

• A. ${(x+a)}^2=\frac{1}{2}(2y-2b)$
• B. ${(x-a)}^2=\frac{1}{2}(2y-2b)$
• C. ${(x+a)}^2=\frac{1}{4}(2y-2b)$
• D. ${(x-a)}^2=\frac{1}{8}(2y-2b)$

Answer 1

Answer: (B)

${(x-a)}^2=\frac{1}{2}(2y-2b)$

Given:

Vertex $=(a,b)$

Length of latus rectum$\left(4a\right)=1$

Axis of parabola = positive y-axis

So we draw the diagram of the parabola as shown in the figure

Now the equation of this parabola would be:

${(x-a)}^2=4a(y-b)$

Putting the value of latus rectum in the equation we get,

${(x-a)}^2=(y-b)$

${(x-a)}^2=\frac{1}{2}(2y-2b)$

This is the required answer.

Option B is the answer