JEE Main 30 June 2022 Shift 1 Solved Paper

Given ellipse x2+a2y2=25a2⇒‌

x2

25a2

+‌

y2

25

=1
‌ eccentricity ‌(e1)=√1−‌

b2

a2

=√1−‌

25

25a2

=√1−‌

1

a2

⇒e12=1−‌

1

a2

Also, given hyperbola,
x2−a2y2=5
⇒‌

x2

5

−‌

y2

5 a2

=1
‌ eccentricity ‌(e2)=√1+‌

b2

a2

=√1+‌

5

5a2

=√1+‌

1

a2

⇒e22=1+‌

1

a2

Also given,
e1=b×e2
⇒e12=b2×e22
⇒1−‌

1

a2

=b2(1+‌

1

a2

)
⇒‌

a2−1

a2

=‌

b2(a2+1)

a2

⇒b2=‌

a2−1

a2+1

y=logex is inverse of y=ex so it is mirror image of each other with respect to y=x line.
Slope of tangent to y=ex curve
‌

dy

dx

=ex
Slope of tangent to y=logex curve,
‌

dy

dx

=‌

1

x

Both tangents are parallel to y=x line for minimum distance condition.
∴ Slope of y=x line = Slope of both the tangent.
∴‌

dy

dx

=ex=1⇒ex=e0=x=0
∴y=ex=e0=1
and ‌

dy

dx

=‌

1

x

=1⇒x=1
∴y=loge1=0
∴ tangent at (0,1) point of y=ex curve and tangent at (1,0) point of y=logex curve are parallel.
∴ Minimum distance between point (0,1) and (1,0) is =√12+12=√2
∴a=√2
So, b2=‌

a2−1

a2+1

=‌

2−1

2+1

=‌

1

3

∴a2+‌

1

b2

=2+‌

1

1∕3

=5