Question

$P{P}^{{}^{\prime }}$ is a diameter of the ellipse $b^2{x}^2+a^2{y}^2=a^2{b}^2$ such that $P{P}^{{}^{\prime }{}^2}$ is the AM of the squares of the major and minor axes. Then the slope of $P{P}^{{}^{\prime }}$ is

• A. $\frac{b}{a}$
• B. $\frac{a}{b}$
• C. $\frac{\pi }{4}$
• D. $\frac{\pi }{3}$

Answer 1

Answer: (A)

$\frac{b}{a}$

Diameter of ellipse $b^2{x}^2+a^2{y}^2=a^2{b}^2$ is $y=-\frac{b^{2}x}{a^{2}m}$

On substituting $y$ in the equation of ellipse, we get

$b^2{x}^2+a^2{(-\frac{b^{2}x}{a^{2}m})}^2=a^2{b}^2\Rightarrow b^2{x}^2+\frac{b^4{x}^2}{a^2{m}^2}=a^2{b}^2\Rightarrow x^2+\frac{b^2{x}^2}{a^2{m}^2}=a^2\Rightarrow (a^2{m}^2+b^2)x^2=a^4{m}^2\Rightarrow x=\pm \frac{a^{2}m}{\sqrt{a^2{m}^2+b^2}}$

We know $y=-\frac{b^{2}x}{a^{2}m}v$

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

$\Rightarrow y=\mp \frac{b^2}{\sqrt{a^2{m}^2+b^2}}$

So, the coordinates of $P$ are $(\frac{a^{2}m}{\sqrt{a^2{m}^2+b^2}},-\frac{b^2}{\sqrt{a^2{m}^2+b^2}})$ and $P^{\prime }$ are $(-\frac{a^{2}m}{\sqrt{a^2{m}^2+b^2}},\frac{b^2}{\sqrt{a^2{m}^2+b^2}})$

$\Rightarrow {P{P}^{\prime }}^2=\frac{4a^4{m}^2}{a^2{m}^2+b^2}+\frac{4b^4}{a^2{m}^2+b^2}\Rightarrow \frac{{4a}^2+4b^2}{2}=\frac{4a^4{m}^2}{a^2{m}^2+b^2}+\frac{4b^4}{a^2{m}^2+b^2}\Rightarrow \frac{a^2+b^2}{2}=\frac{a^4{m}^2+b^4}{a^2{m}^2+b^2}\Rightarrow a^4{m}^2+a^2{m}^2{b}^2+a^2{b}^2+b^4={2a}^4{m}^2+2b^4\Rightarrow (a^4-a^2{b}^2)m^2=a^2{b}^2-b^4\Rightarrow m^2=\frac{b^2(a^2-b^2)}{a^2(a^2-b^2)}\Rightarrow m=\frac{b}{a}$

So, option A is correct.