This Note is for Those who love to use Properties and some innovative Techniques while Solving Question of Conics Section ! These Properties are Highly reduce our Calculation and are very useful sometimes , Specially when we have Time Constrained !

**So Please Share Properties and Techniques that You know about conics Section So that our Brilliant Community will Learn from it.**

So Now Here I Shared some few Techniques ( Properties ) of Conics Section **That are Created By Me** :)

**For Ellipse** ( By Me )

\(\\ \cfrac { { x }^{ 2 } }{ { a }^{ 2 } } +\cfrac { { y }^{ 2 } }{ { b }^{ 2 } } =1\quad \quad \quad (:\quad a>b\quad )\).

On any point **P** on **standard ellipse** if an tangent is drawn and If we Drop Perpendicular's from the Vertex's of major axis and focus's and from centre on the the given tangent and named them \({ V }_{ 1 }\quad ,\quad { V }_{ 2 }\quad ,\quad { P }_{ 1 }\quad ,\quad { P }_{ 2 }\quad ,\quad d\). suitably Then :

1)- \[\displaystyle{ V }_{ 1 }\quad ,\quad d\quad ,\quad { V }_{ 2 }\quad \longrightarrow \quad AP\\ \\ 2d\quad =\quad { V }_{ 1 }+{ V }_{ 2 }\quad \quad \]

2)- \[\displaystyle{ P }_{ 1 }\quad ,\quad d\quad ,\quad { P }_{ 2 }\quad \longrightarrow \quad AP\\ \\ 2d\quad ={ \quad P }_{ 1 }+{ P }_{ 2 }\]

3)- \(\cfrac { { S }_{ 1 }P }{ { S }_{ 2 }P } \quad =\quad \cfrac { \quad P_{ 1 } }{ { \quad P }_{ 2 } } \).

**Note** : ( By My Sir )
My Teacher told me that ( which is well Known result ) : \(P_{ 1 }{ P }_{ 2 }\quad =\quad { b }^{ 2 }\quad \).

**My Turn is Over ! Now it's Your Turn** ,

So please Post Properties or techniques Related To conics Section That are created by You or may also be you learnt it Somewhere else :)

**Reshare** This More and More So that it can reaches to everyone , So that we can Learn from Them!

## Comments

Sort by:

TopNewestThere are actually many properties, here are few of them

If the normal at any point P on the ellipse with centre C meet the major & minor axes in G & g respectively & if CF be perpendicular upon this normal then

PF.PG = \(b^{2}\)

PF.Pg = \(a^{2}\)

PG.Pg = SP.S'P

CG.CT = \(CS^{2}\)

If tangent at the point P of a standard ellipse meets the axes at T & t and CY is perpendicular on it from centre then

Tt.PY = \(a^{2} - b^{2}\)

least value of Tt is \(a + b\)

This is only for ellipse but there are many for each curve – Krishna Sharma · 2 years, 7 months ago

Log in to reply

– Niranjan Khanderia · 2 years, 7 months ago

It would be much helpful if a diagram is drawn. Thanks.Log in to reply

Pleasealso Share it with us , Thanks Krishna :) – Deepanshu Gupta · 2 years, 7 months agoLog in to reply

extremities of latus rectumI don't know how it works but it has worked till now in every problem, I'll try to prove if I got some time meanwhile you can apply and try to prove it :)

Note:- I am not 100% sure with it because I haven't found any case in which this doesn't occur – Krishna Sharma · 2 years, 7 months ago

Log in to reply

Hey, few more, for ellipse:

The portion of the tangent to the ellipse between the point of contact and the directrix subtend a right angle at the corresponding focus.

The circle on the focal distance as diameter touches the auxiliary circle.

Tangent at the extremeties of latus rectum pass through the corresponding foot of directrix on the major axis.

Ratio of area of any triangle inscribed in a standard ellipse ( a> b) and that of triangle formed by corresponding points on the auxiliary circle is \[\frac {b} {a}\]

Log in to reply

– Niranjan Khanderia · 2 years, 7 months ago

It would be much helpful if a diagram is drawn. Thanks.Log in to reply

– Deepanshu Gupta · 2 years, 7 months ago

Thanks alot :)Log in to reply

These are good properties of conic sections. Could you add them to the corresponding Conic Section wiki pages? Thanks! – Calvin Lin Staff · 2 years, 7 months ago

Log in to reply

I have posted my first challenge problem on conic sections. Please post solutions.

Link: https://brilliant.org/problems/conic-sections-challenge-1/?group=Ouxw5LvrW3MT&ref_id=590148

Check out my profile page to get other conic challenges. – Ninad Akolekar · 2 years, 6 months ago

Log in to reply