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Common Tangent to Two Symmetric Ellipses | JEE

JEE Maths question with a full step-by-step solution.

Question
Which one of the following is the common tangent to the ellipses x2a2+b2+y2b2=1\dfrac{x^2}{a^2+b^2}+\dfrac{y^2}{b^2}=1 and x2a2+y2a2+b2=1\dfrac{x^2}{a^2}+\dfrac{y^2}{a^2+b^2}=1?
Aay=bx+a4−a2b2+b4ay=bx+\sqrt{a^4-a^2b^2+b^4}
Bby=ax−a4+a2b2+b4by=ax-\sqrt{a^4+a^2b^2+b^4}correct
Cay=bx−a4+a2b2+b4ay=bx-\sqrt{a^4+a^2b^2+b^4}
Dby=ax+a4−a2b2+b4by=ax+\sqrt{a^4-a^2b^2+b^4}
Solution
Step 1: A tangent of slope mm to the first ellipse is y=mx±(a2+b2)m2+b2y=mx\pm\sqrt{(a^2+b^2)m^2+b^2}. Step 2: For it to also touch the second ellipse, c2=a2m2+(a2+b2)c^2=a^2m^2+(a^2+b^2), so
(a2+b2)m2+b2=a2m2+(a2+b2)  ⇒  b2m2=a2  ⇒  m=±ab.(a^2+b^2)m^2+b^2=a^2m^2+(a^2+b^2)\;\Rightarrow\; b^2m^2=a^2 \;\Rightarrow\; m=\pm\frac{a}{b}.
 Step 3: Take m=abm=\dfrac{a}{b}. Then
c2=(a2+b2)a2b2+b2=a4+a2b2+b4b2.c^2=(a^2+b^2)\frac{a^2}{b^2}+b^2=\frac{a^4+a^2b^2+b^4}{b^2}.
 Step 4: The tangent is y=abx±a4+a2b2+b4by=\dfrac{a}{b}x\pm\dfrac{\sqrt{a^4+a^2b^2+b^4}}{b}, i.e.
by=ax±a4+a2b2+b4.by=ax\pm\sqrt{a^4+a^2b^2+b^4}.
 The matching option is by=ax−a4+a2b2+b4by=ax-\sqrt{a^4+a^2b^2+b^4}. Correct answer: (2)
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