For the hyperbola H : x 2 - y 2 = 1 and the ellipse E : x 2 a 2 + y 2 b 2 = 1 , a > b > 0 , let the…

For the hyperbola H:x2-y2=1 and the ellipse E:x2a2+y2b2=1,a>b>0, let the

(1) eccentricity of E be reciprocal of the eccentricity of H, and

(2) the line y=52x+K be a common tangent of E and H.

Then 4a2+b2 is equal to

Solution

Given H:x2-y2=1,E:x2a2+y2b2=1

eH=2  & eE=1eH=12

For hyperbola  e2=1-b2a2=12b2a2=12

Also given that the common tangent of H & E is y=52x+k i.e.m=52

We know that the condition for common tangent of ellipse x2a2+y2b2=1 and hyperbola x2A2-y2B2=1 is a2m2+b2=A2m2-B2

Now, for common tangency: 52a2+b2=52-1

52+b2a2=32a2a2=12

  a2+b2=12+14=34

 4a2+b2=3

Asked in: JEE Main 2022 (28 Jul Shift 1)

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