select the correct option from the given alternatives: the area of the triangle formed by the line joining the vertex of the parabola x2 = 12y to the endpoints of its latus rectum is \_\_\_\_\_\_\_\_\_ - conic sections - parabola
select the correct option from the given alternatives: the area of the triangle formed by the line joining the vertex of the parabola x2 = 12y to the endpoints of its latus rectum is \_\_\_\_\_\_\_\_\_
18 sq.units explanation: x2 = 12y ∴ 4b = 12 ∴ b = 3 area of triangle = `1/2 xx "ab" xx "os"` = `1/2 xx 4"a" xx "a"` = `1/2(12)(3)` = 18 sq. units
the given equations arex2 + y2 = 16 …(i)and y2 = 6x …(ii)on solving the equation (i) and (ii),or x2 + 6x = 16or x2 + 6x – 16 = 0or (x + 8)(x – 2) = 0or x = 2 or – 8 is not possible solution∴ when x = 2, y = ± = ± = ±equation (i) is a circle with centre (0, 0) and meets axes at (±4,0), (0,±4)equation (ii) represents a parabola with axis as x - axis.points of intersection are a () and c (2,)these are shown in the graph below:area bounded by the circle and parabola= 2[area (oado) + area (adba)]area of circle = π(r) 2= π(4)2= 16π sq. unitsthus, required areathe area of the circle x2 + y2 = 16 which is exterior the parabola y2 = 6x is given the boundaries of the area to be found are,• the curve • the x-axis• x = 0 (y-axis)• x = 4 (a line parallel toy-axis)as per the given boundaries,• the curve,is a curve with vertex at .• x=2 is parallel toy-axis at 2 units away from the y-axis.• x=0 is the y-axis.• the four boundaries of the region to be found are,•point a, where the curve y2 = 6x + 4 and x=0 meet.•point b, where the curve y2 = 6x + 4 and x=2 meet.•point c, where the x-axis and x=2 meet i.e. c(2,0).•point o, or the origin i.e. o(0,0).area of the required region = area of oabc.[using the formula ]the area of the required region solution: equation of parabola `y^(2)=8x` . . .(1)   
 comparing with `y^(2)=4ax`   
 4a=8   
 `rarr" "a=2`   
 co-ordinates of ends of focal chord of parabola (1)   
 `=(at^(2),2at)and((a)/(t^(2)),(-2a)/(t))`   
 `=(2t^(2),4t)and((2)/(t^(2)),(-4)/(t))`   
 the mid-point of this chord is (2,0).   
 `:." "(2t^(2)+(2)/(t^(2)))/(2)=2and(4t-(4)/(t))/(2)=0`   
 `rarr" "t^(2)+(1)/(t^(2))=andt-(1)/(t)=0`   
 `rarr" "t=1`   
 therefore, the co-ordinates of the ends of latus rectum=(2,4) and (2,-4)   
 `:.` equation of latus rectum   
 `y-4=(-4-4)/(2-2)(x-2)`   
 `rarr" "x-2=0`. you don't say that the area must be a rectangle. the true largest area would be a circle, with the circumference equal to 400m and the area would be 12,732.4 m^2. however, evidently, the area is restricted to a rectangle, in which case the maximum area is a square, and the side of the square will be 
400/4 = 100 m. the area is 100x100 = 10,000 m^2.
algebraically,
let x = one pair of opposite sides
(400-2x)/2 = 200-x = the other pair of opposite sides

area = a(x) = x(200 - x) = -x^2 + 200x
x-coordinate of vertex = -b/2a = 200/2 = 100
a(100) = 10,000

that's it! ;) area of trapezium = area of trapezium = = therefore = 24 cm given: (i) xy = 4.5cm

(ii) yz = 5 cm

(iii) zx = 6 cm

to construct: a xyz

steps for construction

step 1 : first we will draw a line segment xy of length 4.5cm



step 2 : we have to draw an arc , taking x as center and radius 6cm



step 3 : we have to another arc, with radius 5cm and center at y and mark the intersection of arcs as z.



step 4 : we have to join xz and xy



this is the required triangle given;y-axis, y = cos x and y = sin x, 0 ≤ x ≤ π/2⇒ sin x = cos xrequired area= (√2 − 1) sq.units let abc be the trianglewhere ab=(a-1)2 cmbc=2√a cmca=(a+1) cmab2=(a-1)2=a2+1-2abc2=(2√a)2=4a2ca2=(a+1)2=a2+1+2ahence ab2+bc2=ac2so Δabc is a right angles triangle at b given:volume of given cube = 64cm3side = a cmvolume = (a)364 = (a)3a = 4cmtwo cubes are joined together.length of resultant cube = 4+4 = 8cmbreadth = 4cmheight = 4cm(a) lsa of cuboid = 2(lh+hb)= 2(8(4) + 4(4))= 2(32+16)= 2(48) = 96cm2⇒ lsa of cuboid = 66m2(b)tsa of cuboid = 2(lh+hb+lb)= 2(8(4) + 4(4) + 8(4))= 2(32+16+32)⇒ tsa of cuboid = 2(80) = 160m2 in this we are asked to divide it in 2:3 ratioso, 3+2 = 55, number congruent of line segment is formed.i. firstly, we have to draw a line segment ab of length 7 cm. draw a line with an acute angle from point aii. point the compass on a, and cut the new line with some radiusplace the compass along the line and make 5 arcs. since we need 5 congruent parts, and name the last one as “g”iii. take length of bg as radius draw an arc from a just below it.again, take the length of ag as radius and draw an arc which cuts the previous arc from b, name the intersection point as hconnect b and hiv. taking the same length that we used in line ag, draw arcs along bh and name the points.v. connect the points along ag and bh. name the intercepts on the line ab as m, n, o & p.vi. thus we can divide a line segment as ao:ob = 6:4= 3:2