Q.The equation of the ellipse whose focus is $(1, -1)$, the directrix the line $x - y - 3 = 0$ and eccentricity $\dfrac{1}{2}$ is
(A) $7x^2 + 2xy + 7y^2 - 10x + 10y + 7 = 0$
(B) $7x^2 + 2xy + 7y^2 + 7 = 0$
(C) $7x^2 + 2xy + 7y^2 + 10x - 10y - 7 = 0$
(D) none

The equation of a conic section is defined by the property that for any point $P(x, y)$ on the conic, the ratio of its distance from the focus $S$ to its distance from the directrix $L$ is a constant, the eccentricity $e$. This is known as the focus-directrix property.

Here, the focus is $S(1, -1)$, the directrix is $L: x - y - 3 = 0$, and the eccentricity $e = \frac{1}{2}$.

1. Calculate the distance $PS$ from $P(x, y)$ to $S(1, -1)$:

   $PS = \sqrt{(x - 1)^2 + (y - (-1))^2} = \sqrt{(x - 1)^2 + (y + 1)^2}$

2. Calculate the perpendicular distance $PM$ from $P(x, y)$ to the directrix $x - y - 3 = 0$:

   $PM = \frac{|x - y - 3|}{\sqrt{1^2 + (-1)^2}} = \frac{|x - y - 3|}{\sqrt{2}}$

3. Substitute these into the focus-directrix formula $PS = e PM$:

   $\sqrt{(x - 1)^2 + (y + 1)^2} = \frac{1}{2} \cdot \frac{|x - y - 3|}{\sqrt{2}}$

   Square both sides to eliminate the square roots:

   $(x - 1)^2 + (y + 1)^2 = \left(\frac{1}{2\sqrt{2}}\right)^2 (x - y - 3)^2$ …

The core concept behind finding the equation of an ellipse (or any conic section) when given its focus, directrix, and eccentricity is the definition of a conic section. This definition states that for any point $P$ on the conic, the ratio of its distance from a fixed point (the focus, $S$) to its perpendicular distance from a fixed line (the directrix, $L$) is a constant value, which is the eccentricity ($e$).

Mathematically, this is expressed as:

$PS = e \cdot PM$

where:

• $P(x, y)$ is any point on the conic.
• $S$ is the focus.
• $L$ is the directrix.
• $PS$ is the distance from point $P$ to the focus $S$.
• $PM$ is the perpendicular distance from point $P$ to the directrix $L$.
• $e$ is the eccentricity.

For an ellipse, the eccentricity $e$ always satisfies $0 < e < 1$. In this problem, $e = \frac{1}{2}$, which confirms we are indeed dealing with an ellipse.

Let's apply this definition step-by-step to find the equation.

1. Identify the given information:

   • Focus $S = (1, -1)$
   • Directrix $L: x - y - 3 = 0$
   • Eccentricity $e = \frac{1}{2}$

2. Let $P(x, y)$ be an arbitrary point on the ellipse.

3. Calculate the distance $PS$ (distance from $P$ to the focus $S$).

   Using the distance formula between two points $(x_1, y_1)$ and $(x_2, y_2)$: $\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$.

   Here, $P(x, y)$ and $S(1, -1)$:

   $PS = \sqrt{(x - 1)^2 + (y - (-1))^2}$

   $PS = \sqrt{(x - 1)^2 + (y + 1)^2}$

4. Calculate the perpendicular distance $PM$ (distance from $P$ to the directrix $L$).

   Using the formula for the perpendicular distance from a point $(x_0, y_0)$ to a line $Ax + By + C = 0$: $\frac{|Ax_0 + By_0 + C|}{\sqrt{A^2 + B^2}}$.

   Here, $P(x, y)$ and $L: x - y - 3 = 0$ (so $A=1, B=-1, C=-3$):

   $PM = \frac{|1 \cdot x + (-1) \cdot y - 3|}{\sqrt{1^2 + (-1)^2}}$

   $PM = \frac{|x - y - 3|}{\sqrt{1 + 1}}$

   $PM = \frac{|x - y - 3|}{\sqrt{2}}$

5. Substitute $PS$, $PM$, and $e$ into the conic section definition $PS = e \cdot PM$.

   $\sqrt{(x - 1)^2 + (y + 1)^2} = \frac{1}{2} \cdot \frac{|x - y - 3|}{\sqrt{2}}$

   $\sqrt{(x - 1)^2 + (y + 1)^2} = \frac{|x - y - 3|}{2\sqrt{2}}$

6. Square both sides of the equation to eliminate the square roots and the absolute value.

   $(x - 1)^2 + (y + 1)^2 = \left(\frac{x - y - 3}{2\sqrt{2}}\right)^2$

   $(x - 1)^2 + (y + 1)^2 = \frac{(x - y - 3)^2}{(2\sqrt{2})^2}$

   $(x - 1)^2 + (y + 1)^2 = \frac{(x - y - 3)^2}{8}$

7. Expand and simplify the equation.

   First, expand the terms on the left side:

   $(x^2 - 2x + 1) + (y^2 + 2y + 1) = \frac{(x - y - 3)^2}{8}$

   $x^2 + y^2 - 2x + 2y + 2 = \frac{(x - y - 3)^2}{8}$ …