JEE Mains 28-July-2022 Shift 1 Solved Paper

The equation of tangent to hyperbola x2−y2=1 within slope m is equal to y=mx±√m2−1.
And for same slope m, equation of tangent to ellipse ‌

x2

a2

+‌

y2

b2

=1 is y=mx±√a2m2+b2....... (ii)
∵ Equation (i) and (ii) are identical
∴a2m2+b2=m2−1
∴m2=‌

1+b2

1−a2

But equation of common tangent is y=√‌

5

2

x+k
∴m=√‌

5

2

⇒‌

5

2

=‌

1+b2

1−a2

∴5a2+2b2=3......‌ (i) ‌
‌ eccentricity of ellipse ‌=‌

1

√2

∴1−‌

b2

a2

=‌

1

2

⇒a2=2b2........‌ (ii) ‌
From equation (i) and (ii) : a2=‌

1

2

,b2=‌

1

4

∴4(a2+b2)=3