Question

Find the equation of the ellipse whose focus is (1,0) the directrix is x+y+1=0 and eccentricity is 1/√2

Answer 1

$\large\underline{\sf{Solution-}}$

Let S(1, 0) be the focus and AB represents the directrix.

Let assume that P(x, y) be any point on the ellipse.

Let PM be the perpendicular drawn from point P on the directrix intersecting directrix at M.

So, By definition of ellipse, we have

$\rm \: SP = e \: PM \: where \: e \: is \: called \: eccentricity$

where,

SP is the distance between focus and point P

PM is perpendicular distance from point P on directrix AB.

Now,

$\rm \: SP^{2} = e ^{2} \: PM^{2}$

$\rm \: {(x - 1)}^{2} + {(y - 0)}^{2} = \dfrac{1}{2} {\bigg |\dfrac{x + y + 1}{ \sqrt{ {1}^{2} + {1}^{2} } } \bigg| }^{2}$

$\rm \: {(x - 1)}^{2} + {y}^{2} = \dfrac{1}{2} {\bigg |\dfrac{x + y + 1}{ \sqrt{ 1 + 1 } } \bigg| }^{2}$

$\rm \: {(x - 1)}^{2} + {y}^{2} = \dfrac{1}{2} {\bigg |\dfrac{x + y + 1}{ \sqrt{2} } \bigg| }^{2}$

$\rm \: {(x - 1)}^{2} + {y}^{2} = \dfrac{1}{4} {(x + y + 1)}^{2}$

$\rm \: 4( {x}^{2} + 1 - 2x + {y}^{2}) = {x}^{2} + {y}^{2} + 1 + 2xy + 2y + 2x$

$\rm \: 4{x}^{2} +4 - 8x + 4{y}^{2} = {x}^{2} + {y}^{2} + 1 + 2xy + 2y + 2x$

$\rm \: 4{x}^{2} +4 - 8x + 4{y}^{2} - {x}^{2} - {y}^{2} - 1 - 2xy - 2y - 2x = 0$

$\rm\implies \:\boxed{\tt{ {3x}^{2} + {3y}^{2} - 2xy - 10x - 2y + 3 = 0}}$

is the required equation of ellipse.

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FORMULA USED

1. Distance Formula

Let A(x₁, y₁) and B(x₂, y₂) be two points in the cartesian plane, then distance between A and B is given by

$\begin{gathered}\boxed{\tt{ AB \: = \sqrt{ {(x_{1} - x_{2}) }^{2} + {(y_{2} - y_{1})}^{2} }}} \\ \end{gathered}$

2. Distance between point and line

Let (p, q) be any point in plane and ax + by + c = 0 be any line in the plane, then the perpendicular distance (d) between point and line is given by

$\boxed{\tt{ \: d \: = \: \bigg |\dfrac{ap + bq + c}{ \sqrt{ {a}^{2} + {b}^{2} } } \bigg| \: }}$

3. Identity

$\boxed{\tt{ {(x + y + z)}^{2} = {x}^{2} + {y}^{2} + {z}^{2} + 2xy + 2yz + 2zx \: }}$