JEE Mains 02-Sep-2020 Shift 2 Solved Paper

Given θ∈(0,

π

2

)
equation of hyperbola ⇒x2− y2sec2θ=10
⇒

x2

10

−

y2

10cos2θ

=1
Hence eccentricity of hyperbola
(eH)=√1+

10cos2θ

10

.....(1) {e=√1+

b2

a2

}
Now equation of ellipse ⇒x2sec2θ+y2=5
⇒

x2

5cos2θ

+

y2

5

=1 {e=√1−

a2

b2

}
Hence eccenticity of ellipse
(e E)=√1−

5cos2θ

5

(e E)=√1−cos2θ =|sin‌θ|=sin‌θ …(2) { ∵θ∈(0,

π

2

)}
given ⇒eH=√5ee
Hence 1+cos2θ=5sin2θ
1+cos2θ=5(1−cos2θ)
1+cos2θ=5−5cos2θ
6cos2θ=4
cos2θ=

2

3

…(3)
Now length of latus rectum of ellipse
=

2a2

b

=

10cos2θ

√5

=

20

3√5

=

4√5

3