Find the area of an equilateral triangle inscribed in x² = 24y whose vertex is at the vertex of the parabola
Answers
Answer: The given parabola is x
2
=12y
On comparing this equation with x
2
=4ay, we get
⇒4a=12
⇒a=3
⇒So, the coordinates of focus is S(0,a)=S(0,3)
⇒Let AB be the latus rectum of the given parabola.
⇒Coordinates of end-points of latus rectum are (−2a,a),(2a,a)
⇒Coordinates of A are (−6,3) while the coordinates of B are (6,3)
Therefore, the vertices of are △OAB are O(0,0),A(−6,3) and B(6,3)
⇒Let (x
1
,y
1
),(x
2
,y
2
),(x
3
,y
3
) be the coordinates of ΔOAB
⇒Area of △ OAB
=
2
1
∣x
1
(y
2
−y
3
)+x
2
(y
3
−y
2
)+x
3
(y
1
−y
2
)∣
=
2
1
∣0(3−3)+(−6)(3−0)+6(0−3)∣
=
2
1
∣(−6)(3)+6(−3)∣
=
2
1
∣−18−18∣
=
2
1
∣−36∣
=
2
1
×36
=18 sq.units
Thus the required area of the triangle is 18 unit
2
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Class 11
>>Maths
>>Conic Sections
>>Parabola
>>Find the area of the triangle formed by
Question
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Find the area of the triangle formed by the lines joining the vertex of the parabola x
2
=12y to the ends of its latus rectum.
Medium
Solution
verified
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The given parabola is x
2
=12y
On comparing this equation with x
2
=4ay, we get
⇒4a=12
⇒a=3
⇒So, the coordinates of focus is S(0,a)=S(0,3)
⇒Let AB be the latus rectum of the given parabola.
⇒Coordinates of end-points of latus rectum are (−2a,a),(2a,a)
⇒Coordinates of A are (−6,3) while the coordinates of B are (6,3)
Therefore, the vertices of are △OAB are O(0,0),A(−6,3) and B(6,3)
⇒Let (x
1
,y
1
),(x
2
,y
2
),(x
3
,y
3
) be the coordinates of ΔOAB
⇒Area of △ OAB
=
2
1
∣x
1
(y
2
−y
3
)+x
2
(y
3
−y
2
)+x
3
(y
1
−y
2
)∣
=
2
1
∣0(3−3)+(−6)(3−0)+6(0−3)∣
=
2
1
∣(−6)(3)+6(−3)∣
=
2
1
∣−18−18∣
=
2
1
∣−36∣
=
2
1
×36
=18 sq.units
Thus the required area of the triangle is 18 unit
2
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