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```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.

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