JEE Mains 06-Sep-2020 Shift 1 Solved Paper

Let foot of perpendicular is (h, k)

x2

4

+

y2

2

=1 (Given)
a=2,b=√2, e=√1−

2

4

=

1

√2

∴ Focus (ae,0)=(√2,0)
Equation of tangent,
y=mx+√a2m2+b2
y=mx+√4m2+2
Passes through (h, k)
(k−mh)2=4m2+2 ....(1)
line perpendicular to tangent will have slope −

1

m

y−0=−

1

m

(x−√2)
my=−x+√2
(h+mk)2=2.....(2)
Add equation (1) and (2)
k2(1+m2)+h2(1+m2)=4(1+m2)
h2+k2=4
x2+y2=4 (Auxiliary circle)
∴(−1,√3) lies on the locus.