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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

\dfrac{x^2}{a^2+b^2}+\dfrac{y^2}{b^2}=1

and

\dfrac{x^2}{a^2}+\dfrac{y^2}{a^2+b^2}=1

ay=bx+\sqrt{a^4-a^2b^2+b^4}

by=ax-\sqrt{a^4+a^2b^2+b^4}

correct

ay=bx-\sqrt{a^4+a^2b^2+b^4}

by=ax+\sqrt{a^4-a^2b^2+b^4}

Solution

Step 1: A tangent of slope

m

to the first ellipse is

y=mx\pm\sqrt{(a^2+b^2)m^2+b^2}

. Step 2: For it to also touch the second ellipse,

c^2=a^2m^2+(a^2+b^2)

, so

(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=\dfrac{a}{b}

. Then

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=\dfrac{a}{b}x\pm\dfrac{\sqrt{a^4+a^2b^2+b^4}}{b}

, i.e.

by=ax\pm\sqrt{a^4+a^2b^2+b^4}.

The matching option is

by=ax-\sqrt{a^4+a^2b^2+b^4}

. Correct answer: (2)

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