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Question

​​​​​​Column IColumn IIa. If x,yR, satisfying the equation (x4)24+y29=1 p. 23 then the difference between the largest and smallest value of the expression x24+y29 is b. If PQ is focal chord of ellipse x225+y216=1 which passes q. 10through S(3,0) and PS=2, then length of chord PQ isc. If the normal at the point P(θ) to the ellipsex214+y25=1 intersect it again at the point Q(2θ), then the value of cosθ is r. 347d. The length of common tangent to x2+y2=16 and s. 89x2+25y2=225 is

A
as; br; cq; dp
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B
ap; bq; cr; ds
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C
as; bq; cp; dr
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D
aq; br; cs; dp
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Solution

The correct option is C as; bq; cp; dr

(a).
Let x4=2 cos θ,
x=2 cos θ+4 and y=3 sin θ
Now,
E=x24+y29
=(2 cos θ+4)24+sin2 θ
=4 cos2 θ+16+16 cos θ+4 sin2 θ4
=20+16 cos θ4
=5+4 cos θ
Hence,
EmaxEmin=91=8

(b).
We know that
1SP+1SQ=2ab2
12+1SQ=1016
SQ=8
PQ=10

(c).
Let P(θ)=(14 cos θ, 5 sin θ)
Normal at P on the ellipse will be
14xcos θ8ysin θ=9
Q(2θ)(14 cos 2θ, 5 sin 2θ) lies on the given eqn of the normal.
14(14cos 2θ)cos θ5(5sin 2θ)sin θ=9
28 cos2 θ1410 cos2 θ=9 cos θ
cos θ=76, or 23 (cos θ/>1)
cos θ=23

(d).
9x2+25y2=225
x225+y29=1
a=5
b=3
Circle C:x2+y2=16
Eqn of tangent to the ellipse at A(5 cos θ, 3 sin θ) is,
x cos θ5+y sin θ3=1
Since this line is also a tangent to the circle,
1cos2 θ25+sin2 θ9=4
116=cos2θ25+sin2θ9
sin2θ=81256cos2θ=175256
And we know that
AB=S1
=(5 cos θ)2+(3 sin θ)216
=9+16 cos2 θ16
=16×1752567
AB=347

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