Question

If the distance of a point P on an ellipse from its focus (1, 2) is half of the distance of P from its corresponding directrix \( x+y=0 \), then the point of intersection of the given directrix and its major axis, is

• A. \( (2,-2) \)
• B. \( (-\frac{1}{2},\frac{1}{2}) \)
• C. \( (-1,1) \)
• D. \( (\frac{1}{3},-\frac{1}{3}) \)

Hint:

The eccentricity value (\( e = 1/2 \)) given in the text is extra data. The axis direction depends entirely on the orientation of the directrix line and the position of the focus point.

Answer 1

The Correct Option is B

Step 1: Understand the geometry.
The major axis of an ellipse goes through the focus and is perpendicular to the directrix. So we build the major-axis line and then find where it cuts the directrix.
Step 2: Find the slope of the directrix.
The directrix is $x+y=0$, i.e. $y=-x$, so its slope is $-1$.
Step 3: Find the slope of the major axis.
The major axis is perpendicular to the directrix, so its slope $m$ satisfies $m\times(-1)=-1$, giving $m=1$.
Step 4: Write the major-axis line.
It passes through the focus $(1,2)$ with slope $1$: \[ y-2=1(x-1)\implies x-y+1=0 \]
Step 5: Solve with the directrix.
Use $y=-x$ in $x-y+1=0$: \[ x-(-x)+1=0\implies 2x+1=0\implies x=-\tfrac{1}{2} \]
Step 6: Find $y$ and conclude.
Then $y=-x=\tfrac{1}{2}$, so the point is \[ \boxed{\left(-\tfrac{1}{2},\ \tfrac{1}{2}\right)} \]