Question

Let \(x+y=0\) be the equation of the latus rectum of a parabola. Let the axis of this parabola pass through the point (1, 1). If \(x+y-2\sqrt{2}=0\) is the equation of the directrix of the parabola, then its vertex is

• A. \((\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}})\)
• B. \((\sqrt{2},\sqrt{2})\)
• C. (0,0)
• D. \((2\sqrt{2},2\sqrt{2})\)

Hint:

The axis of a parabola always passes through the focus and the vertex, and is perpendicular to both the directrix and the latus rectum.

Answer 1

The Correct Option is B

Step 1: Gather the given lines.
The latus rectum is $x+y=0$ and the directrix is $x+y-2\sqrt2=0$; these are parallel. The axis passes through $(1,1)$ and is perpendicular to both. We want the vertex.
Step 2: Find the axis.
The latus rectum has slope $-1$, so the axis (perpendicular to it) has slope $1$. Through $(1,1)$ this gives $y-1=1(x-1)$, i.e. $y=x$.
Step 3: Locate the focus.
The focus is where the axis meets the latus rectum. Solve $y=x$ with $x+y=0$: that gives $2x=0$, so the focus is $S=(0,0)$.
Step 4: Find where the axis meets the directrix.
Solve $y=x$ with $x+y=2\sqrt2$: that gives $2x=2\sqrt2$, so the point is $Z=(\sqrt2,\sqrt2)$.
Step 5: Use the vertex midpoint property.
The vertex is equidistant from the focus and the directrix along the axis, so it is the midpoint of $S$ and $Z$.
Step 6: Compute the midpoint.
\[ V=\left(\frac{0+\sqrt2}{2},\,\frac{0+\sqrt2}{2}\right)=\left(\frac{\sqrt2}{2},\frac{\sqrt2}{2}\right)=\left(\frac{1}{\sqrt2},\frac{1}{\sqrt2}\right). \] So the vertex is $\left(\tfrac{1}{\sqrt2},\tfrac{1}{\sqrt2}\right)$.
\[ \boxed{\left(\dfrac{1}{\sqrt2},\dfrac{1}{\sqrt2}\right)} \]