find the area of triangle formed by lines joining vertex of parabola
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which triangle?there is no figure
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The given equation of the parabola isx2=12y
Now comparing this with the general equation x2=4ayx2=4ay we get,4a=12⇒a=34a=12⇒a=3∴∴ The coordinates of foci are F(0,a)F(0,a). Let AB be the latus rectum of the given parabola.
At y=3,x2=12(3)y=3,x2=12(3)⇒x2=36⇒x2=36∴x=±6∴x=±6∴∴ The coordinates of A are (-6, 3) and coordinates of B are (6,3)∴∴ The vertices of Δ0ABΔ0AB are0(0,0), A(−6,3)and B(6,3)
Area of the triangle is 1/2[x1(y2-y2)+x2(y3-y1)+x2(y1-y2)]
Area of a triangle ABC, whose coordinates are (x1,y1),(x2,y2),(x3,y3)
Substituting the values we get,area of Δ0AB
=1/2[0(3-3)+(-6)(3-0)+6(0-3)
=1/2[-18-18]
=1/2[-36]
=1/2*-36
= 18 sq.unitsHence the required area of the triangle is 18 sq.units.
PLZ MARK IT AS BRAINLIEST
Now comparing this with the general equation x2=4ayx2=4ay we get,4a=12⇒a=34a=12⇒a=3∴∴ The coordinates of foci are F(0,a)F(0,a). Let AB be the latus rectum of the given parabola.
At y=3,x2=12(3)y=3,x2=12(3)⇒x2=36⇒x2=36∴x=±6∴x=±6∴∴ The coordinates of A are (-6, 3) and coordinates of B are (6,3)∴∴ The vertices of Δ0ABΔ0AB are0(0,0), A(−6,3)and B(6,3)
Area of the triangle is 1/2[x1(y2-y2)+x2(y3-y1)+x2(y1-y2)]
Area of a triangle ABC, whose coordinates are (x1,y1),(x2,y2),(x3,y3)
Substituting the values we get,area of Δ0AB
=1/2[0(3-3)+(-6)(3-0)+6(0-3)
=1/2[-18-18]
=1/2[-36]
=1/2*-36
= 18 sq.unitsHence the required area of the triangle is 18 sq.units.
PLZ MARK IT AS BRAINLIEST
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