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The area of the triangle formed by the lines joining the vertex of the parabola x squared equal to 12 to the ends of its latest rectum is?
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The area of the triangle formed by the lines joining the vertex of the...
X^2 = 12y 
we know general equation is x^2= 4ay then vertex is ( 0, 0) and latus reactum is y = a 

compare , x^2 = 12y = 4(3)y 
and x^2 = 4ay 
then, 
a = 3 
so, eqn of Latus rectum ; y = 3 
put y = 3 in eqn of parabola , x^2 = 12y 

x^2 = 36 
x = +- 6 
hence, ( 6, 3) and ( -6 ,3) are the points lies on line and latus rectum .

now , a/ c to question , 
area of ∆ is formed by ( 0, 0) , ( 6, 3) and ( - 6, 3) 

use co- ordinate formula for finding area of ∆ .
area of ∆ = 1/2{ 0( 3 - 3) + 6( 3 - 0) - 6( 0-3) } 

= 1/2{ 0 + 18 + 18 } 

= 18 square unit 

hence, area of ∆ = 18 square unit .

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The area of the triangle formed by the lines joining the vertex of the parabola x squared equal to 12 to the ends of its latest rectum is?
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