Ellipse : IIT -JEE Comprehension Type?
Two tangents PA and PB are drawn from a point P(h,k) to the ellipse x^2/a^2 + y^2/b^2 = 1 (a>b) .Angle of the tangents with the positive x - axis are @1 and @2. Normals at A and B are intersecting at Q point .
Q1. Find the locus of P, if tan@1* tan@2 = 4
Q2.Circumcentre Of triangle QAB is :
A.mid point of AB
B.mid point of PQ
C. orthocentre of triangle PAB
D. can't say
Q3. Locus of P, if cot@1 + cot@2 = lamda( i.e. constant)
Please Explain Your Answer in all the three questions
Update:10 points...........
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Verified answer
Q1
A tangent of gradient m to ellipse satisfies (y−mx)²=b²+a²m² … (standard result)
If P lies in this tangent (k−mh)²=b²+a²m² … (i)
Hence if grad(PA)=m1 and grad(PB)=m2 then m1 and m2 satisfy (i)
But (i) is m²(h²−a²)−2hkm+(k²−b²)=0 … (ii) → m1m2=(k²−b²)/(h²−a²)
Hence (k²−b²)/(h²−a²)=4 and locus is 4x²−y²=4a²−b² … (hyperbola)
Q2
∠PAQ=∠PBQ=90° so PAQB is cyclic because opp ∠s are supplementary. Also, because they are both 90° PQ is a diameter and so the centre of the circum-circle is the mid-point of PQ.
Q3
The locus rule can be written m1+m2=λm1m2.
From (ii) m1+m2=2hk/(h²−a²) and m1m2=(k²−b²)/(h²−a²)
Hence 2hk=λ(k²−b²) and so locus is 2xy=λ(y²−b²) … (hyperbola)
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