Skip to content
dronacharya.ai
← JEE Advanced 2015 · Paper 2
MathematicsInteger+4Question 44 of 60

JEE Advanced 2015 · Paper 2, Mathematics question 44

Suppose that the foci of the ellipse x29+y25=1\dfrac{x^2}9+\dfrac{y^2}5=1 are (f1,0)(f_1,0) and (f2,0)(f_2,0) where f1>0f_1>0 and f2<0f_2<0. Let P1P_1 and P2P_2 be two parabolas with a common vertex at (0,0)(0,0) and with foci at (f1,0)(f_1,0) and (2f2,0)(2f_2,0), respectively. Let T1T_1 be a tangent to P1P_1 which passes through (2f2,0)(2f_2,0) and T2T_2 be a tangent to P2P_2 which passes through (f1,0)(f_1,0). If m1m_1 is the slope of T1T_1 and m2m_2 is the slope of T2T_2, then the value of (1m12+m22)\left(\dfrac1{m_1^2}+m_2^2\right) is

Answer and worked solution

Official answer: 4

Foci. c2=9−5=4c^2=9-5=4, so f1=2f_1=2 and f2=−2f_2=-2.

Parabolas.

  • P1P_1: y2=8xy^2=8x (focus (2, 0)).
  • P2P_2: y2=−16xy^2=-16x (focus (−4, 0)).

Tangent T1T_1 through (−4, 0). Tangents to P1P_1 are y=mx+2my=mx+\frac2m. Then 0=−4m+2m0=-4m+\frac2m, so m12=12m_1^2=\frac12.

Tangent T2T_2 through (2, 0). Tangents to P2P_2 are y=mx−4my=mx-\frac4m. Then 0=2m−4m0=2m-\frac4m, so m22=2m_2^2=2.

Value. 1m12+m22=2+2=4\dfrac1{m_1^2}+m_2^2=2+2=4.

Answer: 4