IIT JEE 2009 Mathematics Paper2 Code 1 Solutions

SINGLE CORRECT CHOICE TYPE

1.     The normal at a point P on the ellipse x² + 4y² = 16 meets the x-axis at Q. If M is the midpoint of the line segment PQ, then the locus of M intersects the latus rectums of the given ellipse at the points

(A) (±(3√5)/2,±2/7) 

(B) (±(3√5)/2,±√19/4) 

(C) (±2√3,±1/7) 

(D) (±2√3,±(4√3)/7)

Sol.   (C)

       Normal to the ellipse is 4xsec f - 2y cosec f = 12

        Q ≡ (3 cosΦ, 0)

        M ≡ (α, β)

          

        α = (3cosΦ + 4cosΦ )/2 = 7/2 cos Φ

        => cos Φ  = 2/7 α

        β = sinΦ

        cos²Φ  + sin²Φ = 1

        => 4/49 α² + β² = 1 => 4/49 x² + y² = 1

        => latus rectum x = + 2√3

        48/49 + y² = 1 => y + 1/7

        Hence , the points are (+2√3, +1/7).

 

2.     The locus of the orthocenter of the triangle formed by the lines 
(1+p)x - py + p(1 + p) = 0, (1 + q)x - qy + q(1 + q) = 0 and y = 0, where p ≠ q, is

        (A)    a hyperbola

        (B)    a parabola

        (C)    an ellipse

        (D)    a straight line

Sol.   (D)

        Intersection point of y = 0 with first line is the point B(-p, 0)

        Intersection point of y = 0 with second line is the point A(-q, 0)

        Intersection point of the two lines is the point C(pq, (p + 1)(q + 1))

        Altitude from C to AB is x = pq

        Altitude from B to AC is y = -q/1+q (x + p)

        On solving these two we get x = pq and y = -pq

        => locus of orthocenter is x + y = 0.

« Back || Next »