Question

Which of the following is the common tangent to the ellipse $\frac{x^2}{a^2+b^2}+\frac{y^2}{b^2}=1$ & $\frac{x^2}{a^2}+\frac{y^2}{a^2+b^2}=1$?

• A. $ay=bx+\sqrt{a^4-a^2{b}^2+b^4}$
• B. $by=ax-\sqrt{a^4+a^2{b}^2+b^4}$
• C. $ay=bx-\sqrt{a^4+a^2{b}^2+b^4}$
• D. $by=ax-\sqrt{a^4-a^2{b}^2+b^4}$

Answer 1

Answer: (B)

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

$\frac{x^2}{a^2+b^2}+\frac{y^2}{b^2}=1$ and $\frac{x^2}{a^2}+\frac{y^2}{a^2+b^2}=1$

Equation of tangent for first ellipse

$y=mx\pm \sqrt{(a^2+b^2)m^2+b^2}$, where $m$ is slope.

Equation of tangent for second ellipse

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

Equating the both tangential equations we get

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

$⟹m=\pm \frac{a}{b}$

$y=\pm \frac{a}{b}x\pm \sqrt{(a^2+b^2)(\frac{a}{b}{)}^2+b^2}$

$by=\pm ax\pm \sqrt{(a^2+b^2)(a{)}^2+b^4}$

For $m=\frac{a}{b}$

$by=ax\pm \sqrt{(a^2+b^2)(a{)}^2+b^4}$