Let (x2/a2)+(y2/b2)=1(ab) be a given ellipse, length of whose latus rectum is 10 . If its eccentricity is the maximum value of the function, φ( t )=(5/12) +t-t2, then a2+b2 is equal to :

Question

Let (x2/a2)+(y2/b2)=1(a>b) be a given ellipse, length of whose latus rectum is 10 . If its eccentricity is the maximum value of the function, φ( t )=(5/12) +t-t2, then a2+b2 is equal to : (A) 126 (B) 135 (C) 145 (D) 116

Tardigrade Admin · Accepted Answer

(x2/a2)+(y2/b2)=1(a>b) ; (2 b2/a)=10 ⇒ b2=5 a ldots(i) Now, φ(t)=(5/12)+t-t2=(8/12)-(t-(1/2))2 φ(t) max =(8/12)=(2/3)=e ⇒ e2=1-(b2/a2)=(4/9) ldots (ii) ⇒ a2=81 ( from (i) &(i i)) So, a2+b2=81+45=126

Q. Let a2x2​+b2y2​=1(a>b) be a given ellipse, length of whose latus rectum is 10. If its eccentricity is the maximum value of the function, ϕ(t)=125​ +t−t2, then a2+b2 is equal to :

126

135

145

116

a2x2​+b2y2​=1(a>b);a2b2​=10 ⇒b2=5a…(i) Now, ϕ(t)=125​+t−t2=128​−(t−21​)2 ϕ(t)max​=128​=32​=e ⇒e2=1−a2b2​=94​… (ii) ⇒a2=81( from (i)&(ii)) So, a2+b2=81+45=126

Q. Let $\frac{x ^{2}}{a ^{2}} + \frac{y ^{2}}{b ^{2}} = 1 (a > b)$ be a given ellipse, length of whose latus rectum is $10.$ If its eccentricity is the maximum value of the function, $ϕ (t) = \frac{5}{12}$ $+ t - t^{2},$ then $a^{2} + b^{2}$ is equal to :