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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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IIT JEE 2009 Mathematics Paper2 Code 1 Solutions...