Equation of Ellipse

Easy question. The trick is to understand the importance of change of co-ordinate axes.

Firstly equation joining PQ is the major axis. Also mid-point of PQ is the centre of the ellipse. The line passing through the centre and perpendicular to the major axis is the minor axis.

So we write Major axis is $x+y-6=0$ and centre of the ellipse is M=(3,3) and the equation of minor axis is $x=y$

Now it is given that (2,2) is a point on the ellipse. By little observation note that it lies on the minor axis.

The length of semi-minor axis is $b=\sqrt { { (3-2) }^{ 2 }+{ (3-2) }^{ 2 } } =\sqrt { 2 }$

Also distance of focus from the centre is $c=\sqrt { { (2-3) }^{ 2 }+{ (4-3) }^{ 2 } } =\sqrt { 2 }$

And we know ${ a }^{ 2 }-{ b }^{ 2 }={ c }^{ 2 }$

Using this we get $a=2$ (a=Length of semi-major axis)

Here comes the final thing.We write eqaution of ellipse in standard form in cartesian co-ordinates as :

$\frac { { x }^{ 2 } }{ { a }^{ 2 } } +\frac { { y }^{ 2 } }{ { b }^{ 2 } } =1$

Now x-co-ordinate is the perpendicular distance from the y-axis and y-co-ordinate is the perpendicular distance from the x-axis.

Following the similar analogy we can write equation of our ellipse as :

$\frac { { (Perp.\quad distance\quad from\quad x-y=0) }^{ 2 } }{ { a }^{ 2 } } +\frac { { (Perp.\quad distance\quad form\quad x+y-6=0) }^{ 2 } }{ { b }^{ 2 } } =1$

Perpendicular distance from $x-y=0$ is $=\left| \frac { x-y }{ \sqrt { 2 } } \right|$

Perpendicular distance from $x+y-6=0$ is $= \left| \frac { x+y-6 }{ \sqrt { 2 } } \right|$

Finally we write our equation as :

$\frac { { \left| \frac { x-y }{ \sqrt { 2 } } \right| }^{ 2 } }{ { 2 }^{ 2 } } +\frac { { \left| \frac { x+y-6 }{ \sqrt { 2 } } \right| }^{ 2 } }{ { (\sqrt { 2 } })^{ 2 } } =1$

After simplifying we get :

$3{ x }^{ 2 }+3{ y }^{ 2 }+2xy-24x-24y+64=0$

Finally we get $\boxed { a+b+c+d+e+f=3+3+2-24-24+64=24 }$

We Can Define Ellipse In A Bit Different Way.

The Locus Of A Point Which moves such that the sum of its distances from

two fixed points(the focii) always remains constant which is equal to the

length of major axis .

Now the point (2,2) lies on the ellipse therefore finding the sum of its

distances from focii we get the sum equal to 4 .

let (x,y) be any point on the ellipse

Again its Sum of distance from P And Q Will Be Equal To 4 .

Equating the distances and after a bit of some manipulations we get the

required answer

the question says find the maximum value of the following "equation"?? I think it should be "expression". lol :)