Question

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

Answer 1

Hint:When solving these types of problems,first define what is ellipse and then using the appropriate formula in accordance to the question asked ,apply it and solve it.

Complete step by step solution:
Given that:
Let S be the focus of the ellipse and ‘e’ be the eccentricity of the ellipse.
Let us consider that P(x,y) be any point on the ellipse,
Eccentricity, \[e = \dfrac{1}{2}\]
Therefore, according to the definition of ellipse,
   $\begin{array}{l}
SP = ePM\\
 \Rightarrow S{P^2} = {e^2}P{M^2}
\end{array}$
i.e.
$\begin{array}{l}
S{P^2} = \dfrac{1}{4}{(PM)^2}\\
 \Rightarrow 4S{P^2} = P{M^2}
\end{array}$
 Applying distance formula-
$4[{(x + 1)^2} + {(y - 1)^2}] = {(\dfrac{{x - y - 3}}{{\sqrt {{1^2} + {{( - 1)}^2}} }})^2}$
 Applying the formula of
  $\begin{array}{l}
{(a + b)^2} = {a^2} + 2ab + {b^2}\\
{(a - b)^2} = {a^2} - 2ab + {b^2}
\end{array}$
$4[{x^2} + 2x + 1 + {y^2} - 2y + 1] = \dfrac{{{{(x - y + 3)}^2}}}{{{{(\sqrt 2 )}^2}}}$
Doing cross multiplication -
$8[{x^2} + 2x + 1 + {y^2} - 2y + 1] = {(x - y + 3)^2}$
Solving $(x - y + 3)$ separately and substitute in the above equation –
Substitute $x - y = a$
 $\begin{array}{l}
{(x - y + 3)^2} = {(a + 3)^2}\\
\end{array}$
  $\begin{array}{l}
 \Rightarrow {(a + 3)^2} = {a^2} + 6a + 9\\
\end{array}$
 Substitute again $a = x - y$
 ${a^2} + 6a + 9 = {(x - y)^2} + 6(x - y) + 9$
Now, solve we know that ${(x - y)^2} = {x^2} - 2xy + {y^2}$
$\begin{array}{l}
{(x - y - 3)^2} = {x^2} - 2xy + {y^2} + 6(x - y) + 9\\
{(x - y - 3)^2} = {x^2} - 2xy + {y^2} + 6x - 6y + 9
\end{array}$
$8[{x^2} + 2x + 1 + {y^2} - 2y + 1] = {x^2} - 2xy + {y^2} + 6(x - y) + 9$
$\begin{array}{l}
8[{x^2} + 2x + 1 + {y^2} - 2y + 1] = {x^2} - 2xy + {y^2} + 6(x - y) + 9\\
8{x^2} + 16x + 8 + 8{y^2} - 16y + 8 = {x^2} - 2xy + {y^2} + 6x - 6y + 9
\end{array}$
Simplify the above equation-
$\begin{array}{l}
7{x^2} + 10x - 1 + 7{y^2} - 10y + 8 + 2xy = 0\\
7{x^2} + 10x + 7{y^2} - 10y + 2xy + 7 = 0
\end{array}$
Arranging the above equation- $7{x^2} + 2xy + 7{y^2} + 10x - 10y + 7 = 0$
This is the required solution.
Hence, from the given multiple options, option B is the correct answer.
Additional Information: An ellipse is the plane curve surrounding the two focal points such that for all the points on the curve, the sum of the two distances to the focal points is the constant. The elongation of an ellipse is the measure of its eccentricity e and it ranges from $0$ to $1$.

Note: Instead of substitution method and making to the whole square of terms as used in the above example one can use ${(a + b + c)^2} = {a^2} + {b^2} + {c^2} + 2ab + 2bc + 2ca$
When we take any terms from the right hand side to the left hand side the sign of the terms changes i.e. negative changes to positive and positive changes to the negative terms.