Please solve q no 4????
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x225+y216=1middle point is (12,25)Let " T " be the equation of the chord . Let " S " denote the equation of the ellipse . Then the chord having its middle point as (x1,y1) is given by T = S1 Here , T ≡ xx125+yy116 − 1S1= x1225+y1216−1 Put , x1=12,y1=25 . The equation of the chord is :x50+y40=x100+y1004x+5y200=21004x+5y=45y=4−4xy=45(1−x)putting in eqn of ellipse:x225+y216=1x225+(45(1−x))216=1x225+(1−x)225=1x2+1+x2−2x25=12x2−2x+1=252x2−2x−24=0x2−x−12=0x2−4x+3x−12=0x(x−4)+3(x−4)=0(x−4)(x+3)=0hence, x=4 or x=−3for x=4y=45(1−x)=45(1−4)=−125for x=−3y=45(1−x)=45(1+3)=165hence end points are A(4,−125)B(3,165)AB=(3+4)2+(165+125)2−−−−−−−−−−−−−−−−−√=49+78425−−−−−−−√=152009−−−−√
option (b)..7/5×41^2
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