Question

Let L be an end of the latus rectum of y^2=4x. The normal at L meets the curve again at M. The normal at M meets the curve again at N. The area of ΔLMN is ?

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Answer 1

Given :-

• L is an end of the latus rectum of y^2=4x.

• The normal at L meets the curve again at M.

• The normal at M meets the curve again at N.

To find:-

• The area of ΔLMN is ?

Solution:-

Let,coordinates of L&M are :

$\sf L(at_1 ^2 , 2at_1 ) ; M (at_2 ^2 ; 2at_2 ) \ respectively.$

We know :

$\\ \\ \mapsto \sf \: y {}^{2} = 4ax...(1)$

And,

$\\ \\ \mapsto \sf \: {y}^{2} = 4x...(2)$

By comparing eqn (1)&(2) : -

» 4ax = 4x

» a = 1

Now As a=1 , coordinates of L and M will be :

L (t_1 ^2 ,2t_1 ) & M(t_2 ^2 , 2t_2 )

Given that normal is at point N.

$\\ \\ \\ \sf \implies \: t_1 {}^{'} = - t_1 - \frac{2}{t_1} = - 3 \\ \\ \\ \implies \sf \: t_2 {}^{'} = - t_2 - \frac{2}{ t_2 }$

» L = [9a ,-6a] ; M = [-9a,6a]

As a=1 ,

» L = (9, -6 ) ; M = (-9,6)

$\\ \\ \\ \sf \implies \: LM {}^{2} = ( \frac{ {(9 + 9)}^{2} + ( - 6 - 6) {}^{2} }{3} ) \\ \\ \\ \implies \sf \: LM = \frac{1280}{3}$

Therefore , LN = 2/3 .

➪ Area of ∆ LMN = 1/2 × 1280/3 × 2/3

= 1280 /9.

$\therefore \boxed{ \bigstar \sf Area \: of \: \triangle LMN \: = \frac{1280}{9} }$

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