find the equation of that focal chord of the parabola `y^(2)=8x` whose mid-point is (2,0).
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`.
let c1 and c2 are two concentric circles with center o and radius of outer circle is 5 cm.given : ac is a chord with length 8 cm that is tangent to inner circle .to find : radius of inner circle i.e. odac is a tangent for c1 at d so,od ⏊ ac [tangent at a point on the circle is perpendicular to the radius through point of contact ]so, oad is a right-angled triangle at d .also it implies that od is perpendicular to the chord ac in c2so we have,ad = dc [perpendicular from the center to the chord bisects the chord]ac = ad + dc8 = ad + adad = 4 cmin △oad by pythagoras theorem(oa)2 = (od)2 + (ad)2(5)2 = (od)2+ (4)225 = (od)2 + 16(od)2= 9od = 3 cm 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 you seem to be missing some information about e.if e is the midpoint between d and f, thene is (1,1) take half of (4-2) = 1 and half of (5-3) = 1e alphabetically does come between d and f.just average the x values and average the y values of the two points d and for plot them graphically, carefully, maybe on some graph paper, and you'llvisually see the midpoint is at (1,1) if , passing through (1, 2), satisfies the differential equation , then multiply 2 in both lhs and rhs,integrating on both the sides, if the figure below represents a parabola, identity the physical&nbsp;quantities tan a/2 is equal to \[\ce{fecl2 +2naoh -> fe(oh)2 + 2naci}\] (a) f = 1/t = 2.11 hz

(b) 2*pi*f = 13.26 radians/s = sqrt(k/m)
m = 0.296 kg
solve for k
(c) (1/2)ka^2 = 0.150 j
solve for amplitude a, in meters formula :- therefore ,put -2x dx = dt