Question

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

• 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 $by=ax-\sqrt{a^4+a^2{b}^2+b^4}$

Slope of common tangent to $\frac{x^2}{a^2+b^2}+\frac{y^2}{b^2}=1$ and $\frac{x^2}{a^2}+\frac{y^2}{a^2+b^2}=1$ will be same as slope of two ellipse.

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

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

$a^4+b^4+2a^2{b}^2=a^2{b}^2$

$a^4+b^4-a^2{b}^2=0$

$a^4+b^4=a^2{b}^2$

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

$y=\frac{a}{b}x-\frac{\sqrt{a^4+a^2{b}^2+b^4}}{b}$

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

$a^3+{ab}^2=b^3{a}^3=a^{2}b+b^3$

$a^3-b^3={ab}^2{a}^3-b^3=a^{2}b$

So,

${ab}^2=a^{2}b$

and, $a=b$

Thus, $by=ax-\sqrt{a^4+a^2{b}^2+b^4}$ is the correct answer.