JEE Coordinate (1)

Geometry Level 4

An ellipse has foci at $(9,20)$ and $(49,55)$ in the $xy$ plane and is tangent to the $x$ -axis.

If length of its major axis is $A$ , report the answer closest integer to $\sqrt A$ .

This is a part of my set Practice for JEE 2017!

3 9 2 6 5 8 4 7

2 solutions

Indraneel Mukhopadhyaya
Jan 13, 2017

Useful property for ellipse : If S1 and S2 are the two foci of ellipse E, and line L is a tangent to E at some point on it, then reflect any one of the two foci, suppose S1, in line L to get point $S1'$ . Then we have the result that $S1'$ $S2$ = length of major axis of E.

Its a nice property i heard it first time !

I Did by using property that foot of perpendicular to tangent from focus lies on auxiliary circle

Prakhar Bindal - 4 years, 4 months ago

@Prakhar Bindal It is a simple consequence of midpoint theorem.

Indraneel Mukhopadhyaya - 4 years, 4 months ago

This is a very unorthodox solution. Since, it's a JEE practise problem, we have all rights to eliminate options.

Clearly length of major axis is always greater than the distance b/w the focii of the ellipse. The question is to find the square root of the length of the major axis.

Now find the distance b/w the focii using the distance formula. It comes out to be $\sqrt{2825}=53.1$ .

Therefore, length of major axis has to be greater than $53$ (Let's approximate it to 53)

Some numbers like $64$ or $81$ can be fancied, whose square roots are $8$ and $9$ . Hence all the options except $8$ and $9$ can be eliminated!

I marked $8$ , answer is $9$ . :P

Hence, I'm writing this as a comment, as there's no access for writing solution as I got it wrong!

Skanda Prasad - 3 years, 1 month ago

Aniket Sanghi
Jan 16, 2017

Product of perpendicular distances upon any tangents from 2 focii = $b^2$ .

And Distance between focii = 2ae.

JEE Coordinate (1)

This is a part of my set Practice for JEE 2017!

2 solutions

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