Question

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

• 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

The correct option is B $by=ax+\sqrt{a^4+a^2{b}^2+b^4}$

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

$y=mx+c$ is tangent

$c^2=(a^2+b^2)m^2+b^2$ …….$\left(1\right)$

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

$c^2=(a^2+b^2)+m^2{a}^2$ ……..$\left(2\right)$

$(a^2+b^2)m^2+b^2=a^2+b^2+m^2{a}^2$

$\Rightarrow a^2{m}^2+b^2{m}^2+b^2=a^2+b^2+m^2{a}^2$

$\therefore a^2=m^2{b}^2$

$\Rightarrow m^2=\frac{a^2}{b^2}$

$\therefore a=\pm mb$

$\Rightarrow m=\pm \frac{a}{b}$

$c^2=(a^2+b^2)\frac{a^2}{b^2}+b^2$

$c^2=\frac{a^4}{b^2}+a^2+b^2$

$c^2=\frac{a^4+b^4+a^2{b}^2}{b^2}$

$c=\pm \frac{\sqrt{a^4+b^4+a^2{b}^2}}{b}$

$\therefore y=\frac{a}{b}c+\frac{\sqrt{a^4+b^4+a^2{b}^2}}{b}$

$\Rightarrow by=ax+\sqrt{a^4+b^4+a^2{b}^2}$ ……….(i)

$by=-ax+\sqrt{a^4+b^4+a^2{b}^2}$ ……….(ii)

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

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