Triangle formed by tangents to a parabola

  • Thread starter takando12
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Homework Statement


The triangle formed by the tangents to the parabola y2=4ax, at the ends of the latus rectum and the double ordinate through it's focus is:
1) equilateral 2) acute angles isosceles 3)right angled isosceles 4) dependent on value of a

Homework Equations




The Attempt at a Solution


The double ordinate through the focus is the latus rectum. I tried to find the intersection of the two tangents by using the AM and GM of the co-ordinates of the points of contact with the parabola .
the two points of contact : (a,2a) and (a,-2a)
x-coordinate of intersection = GM=√a2=a
y-coordinate of intersection = AM= 2a-2a/2 =0
Which gives me the intersection of the two tangents to be (a,0) which is the co-ordinates of the focus. This is clearly wrong. Where am I going wrong in my approach?
 

Answers and Replies

  • #2
Samy_A
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Maybe you should post how you compute(d) the slope of the tangents at the two points of contact.
 
  • #3
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Homework Statement


The triangle formed by the tangents to the parabola y2=4ax, at the ends of the latus rectum and the double ordinate through it's focus is:
1) equilateral 2) acute angles isosceles 3)right angled isosceles 4) dependent on value of a

Homework Equations




The Attempt at a Solution


The double ordinate through the focus is the latus rectum. I tried to find the intersection of the two tangents by using the AM and GM of the co-ordinates of the points of contact with the parabola .
the two points of contact : (a,2a) and (a,-2a)
x-coordinate of intersection = GM=√a2=a
y-coordinate of intersection = AM= 2a-2a/2 =0
Which gives me the intersection of the two tangents to be (a,0) which is the co-ordinates of the focus. This is clearly wrong. Where am I going wrong in my approach?
Since x = √a2 Clearly a is not equal to +a therefore the the abcissa is -a.
And the option three is correct
 

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  • #4
LCKurtz
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Have you noticed you are replying to a post that was made almost 3 years ago?
 

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