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Question Number 30002 by rahul 19 last updated on 14/Feb/18

Commented by ajfour last updated on 14/Feb/18

wrong options, i think.

wrongoptions,ithink.

Commented by ajfour last updated on 15/Feb/18

Commented by ajfour last updated on 15/Feb/18

let slope of common tangent be m. m=(dy/dx)∣_x_1 = −2x_1 +1 ...(i) y_1 =−x_1 ^2 +x_1 +(1/2) .....(ii) slope of FM=((k−(1/2))/(h−(1/2))) = −(1/m) ⇒ m=((2h−1)/(1−2k)) .....(iii) eq. of tangent y−y_1 =m(x−x_1 ) mid point of FM is (h+(1/2) , k+(1/2)) It lies on tangent, hence k+(1/2)−2y_1 =m(h+(1/2)−2x_1 ) ...(iv) using (i), (ii) and (iii) in (iv) : k+(1/2)+2x_1 ^2 −2x_1 −1= (1−2x_1 )(h+(1/2)−2x_1 ) k+(1/2)+2x_1 ^2 −2x_1 −1=h+(1/2)−2x_1 −2hx_1 −x_1 +4x_1 ^2 k−1=h(1−2x_1 )+2x_1 ^2 −x_1 k−1=(h−x_1 )(1−2x_1 ) k−1=(h−x_1 )m k−1=(h+((m−1)/2))m 2(k−1)=(2h−1+m)m 2(k−1)=(2h−1+((2h−1)/(1−2k)))(((2h−1)/(1−2k))) 2(k−1)=(((2h−1)^2 )/((1−2k)))(((2−2k)/(1−2k))) ⇒ (k−1)[1+(((2h−1)/(1−2k)))^2 ]=0 ⇒ k=1 hence locus is y=1 . (There has to be a shorter way, i couldn′t find presently)

letslopeofcommontangentbem.m=dydx∣x1=−2x1+1...(i)y1=−x12+x1+12.....(ii)slopeofFM=k−12h−12=−1m⇒m=2h−11−2k.....(iii)eq.oftangenty−y1=m(x−x1)midpointofFMis(h+12,k+12)Itliesontangent,hencek+12−2y1=m(h+12−2x1)...(iv)using(i),(ii)and(iii)in(iv):k+12+2x12−2x1−1=(1−2x1)(h+12−2x1)k+12+2x12−2x1−1=h+12−2x1−2hx1−x1+4x12k−1=h(1−2x1)+2x12−x1k−1=(h−x1)(1−2x1)k−1=(h−x1)mk−1=(h+m−12)m2(k−1)=(2h−1+m)m2(k−1)=(2h−1+2h−11−2k)(2h−11−2k)2(k−1)=(2h−1)2(1−2k)(2−2k1−2k)⇒(k−1)[1+(2h−11−2k)2]=0⇒k=1hencelocusisy=1.(Therehastobeashorterway,icouldn′tfindpresently)

Commented by rahul 19 last updated on 15/Feb/18

thank u sir! ωo𝛚.

thankusir!ωoω.

Commented by rahul 19 last updated on 15/Feb/18

ok sir, then plz provide your solution. ignoring the options.

oksir,thenplzprovideyoursolution.ignoringtheoptions.

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