 Click to Chat

1800-1023-196

+91-120-4616500

CART 0

• 0

MY CART (5)

Use Coupon: CART20 and get 20% off on all online Study Material

ITEM
DETAILS
MRP
DISCOUNT
FINAL PRICE
Total Price: Rs.

There are no items in this cart.
Continue Shopping
        what is the proof of ellipse equation? and other proofs like latus rectum, please explain me with details.
7 years ago

							Dear Simhadri,
The derivation of the cartesian form for an ellipse is simple and instructive. One simple definition of the ellipse is the "locus of all points of the plane whose distances to two fixed points(called the foci) add to the same constant". See ellipse for other definitions.
Let the foci be points (-c,0) and (c,0). Then the ellipse center is (0, 0).If (x,y) is any point on the ellipse and if $d_1$ is the distance between (x,y) and (-c,0) and $d_2$ is the distance between (x,y) and (c,0), i.e. then we can define $a$ $d_1 + d_2 = 2a\,$
(a here is the semi-major axis, although this is irrelevant for the sake of the proof). From this simple definition we can derive the cartesian equation. Substituting: $\sqrt {(x+c)^2+y^2} + \sqrt {(x-c)^2+y^2} = 2a$
To simplify we isolate the radical and square both sides. $\sqrt {(x+c)^2+y^2} = 2a - \sqrt {(x-c)^2+y^2}$ $(x+c)^2 + y^2 = \left ( 2a - \sqrt{(x-c)^2+y^2} \right )^2$ $(x+c)^2 + y^2 = 4a^2 - 4a\sqrt{(x-c)^2+y^2} + (x-c)^2 +y^2$
Solving for the root and simplifying: $(x+c)^2 + y^2 - (x-c)^2 - y^2 - 4a^2 = - 4a\sqrt{(x-c)^2+y^2}$ $-{1 \over 4a}((x+c)^2 + y^2 - 4a^2 - (x-c)^2 - y^2) = \sqrt{(x-c)^2+y^2}$ $\sqrt{(x-c)^2+y^2} = -{1 \over 4a} ((x+c)^2+y^2-4a^2-(x-c)^2-y^2)$ $\sqrt{(x-c)^2+y^2} = -{1 \over 4a} (x^2 + 2xc + c^2 -4a^2 -x^2 +2xc -c^2)$ $\sqrt{(x-c)^2+y^2} = -{1 \over 4a} (4xc - 4a^2)$ $\sqrt{(x-c)^2+y^2} = a - {c \over a}x$
A final squaring $(x-c)^2+y^2 = a^2 - 2xc + {c^2 \over a^2}x^2$ $x^2 - 2xc + c^2 + y^2 = a^2 -2xc + {c^2 \over a^2}x^2$ $x^2 + c^2 + y^2 = a^2 + {c^2 \over a^2}x^2$
Grouping the x-terms and dividing by $a^2-c^2\,$ $x^2 - {c^2 \over a^2}x^2 + y^2 = a^2 - c^2$ $x^2 \left( 1 - {c^2 \over a^2} \right) + y^2 = a^2 - c^2$
Where: $1 = {a^2 \over a^2}$ $x^2 \left( {a^2 - c^2 \over a^2} \right) + y^2 = a^2 - c^2$ ${x^2 \over a^2} + {y^2 \over a^2-c^2} = 1$
If x = 0 then $d_1 = d_2 = a = \sqrt {c^2+b^2}$
(where b is the semi-minor axis)
Therefore we can substitute $b^2 = a^2 - c^2\,$
And we have our desired equation: ${x^2 \over a^2} + {y^2 \over b^2} = 1$
Cracking IIT just got more exciting,It s not just all about getting assistance from IITians, alongside Target Achievement and Rewards play an important role. ASKIITIANS has it all for you, wherein you get assistance only from IITians for your preparation and win by answering queries in the discussion forums. Reward points 5 + 15 for all those who upload their pic and download the ASKIITIANS Toolbar, just a simple  to download the toolbar….
So start the brain storming…. become a leader with Elite Expert League ASKIITIANS
Thanks
Aman Bansal

7 years ago
Think You Can Provide A Better Answer ?

## Other Related Questions on Analytical Geometry

View all Questions »  ### Course Features

• 731 Video Lectures
• Revision Notes
• Previous Year Papers
• Mind Map
• Study Planner
• NCERT Solutions
• Discussion Forum
• Test paper with Video Solution  ### Course Features

• 53 Video Lectures
• Revision Notes
• Test paper with Video Solution
• Mind Map
• Study Planner
• NCERT Solutions
• Discussion Forum
• Previous Year Exam Questions