Derivation of Standard Equation of Ellipse Video Lecture | Mathematics (Maths) for JEE Main & Advanced

hello this video is on deriving the
equation of a horizontal ellipse hit the
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is the equation of a horizontal ellipse
but where does it come from
first we need two points called the fo
side let's connect a string and when we
extend the string around the foci we
create an ellipse the length of the
string does not change in other words it
is constant when we extend the string
horizontally it is the longest diameter
of the ellipse called the major axis and
the shortest dimer of the lips is called
the minor axis
let's call half the length of the major
axis a cap the length of the minor axis
B and the distance from the center to
one of the foci is C since half the
length of major axis is a then the whole
length is actually 2a when we extend the
string evenly at the top we want to form
a triangle with sides a B and C by the
Pythagorean theorem these sites have the
relationship of B squared plus C squared
is equal to a squared and moving C
squared to the other side gives us that
B squared is equal to a squared minus C
squared what we want to do is set up an
equation that represents all the points
XY that make up the ellipse let's label
the distances from the XY coordinate to
each fo side D 1 and D 2 placing the
ellipse with the center at the origin we
have the following coordinates for the
foe sign remember that the length of the
string is equal to the length of the
major axis which is equal to 2a
therefore d1 plus d2 is equal to 2a next
let's use the distance formula we want
to find the distance of d1 by
substituting the coordinates into the
formula we will also substitute the
coordinates of d2 into the formula what
we need to do next is a series of
algebra steps in order to simplify our
equation so let's begin after
simplifying the equation we arrange the
terms let's square both sides to get rid
of the square roots
expand the parentheses and cancel out
like terms rearrange the terms again and
cancel out the fours square both sides
again to eliminate the square root and
rearrange the equation
let's factor and we have this result
let's divide both sides by a squared
times a squared minus C squared when we
eliminate the common factors we finally
have x squared over a squared plus y
squared over a squared minus C squared
is equal to 1 remember that we showed
that V squared is equal to a squared
minus C squared therefore when we
substitute B squared into our equation
we have the equation of a horizontal
ellipse thank you for watching this
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