Question

If the normals at $P\left(\theta \right)$ and $Q\left(\frac{\pi }{2}+\theta \right)$ to the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1;a>b$ meet the major axis at $G$ and $g$, respectively, then the value of $P{G}^2+Q{g}^2$ is

• A. $b^2(1-e^2)(2-e^2)$
• B. $a^2(e^2-1)(e^2-2)$
• C. $a^2(1+e^2)(2+e^2)$
• D. $b^2(1+e^2)(2+e^2)$

Answer 1

Answer: (B)

$a^2(e^2-1)(e^2-2)$

$P\left(\theta \right)=\frac{ax}{\cos \theta }-\frac{by}{\sin \theta }=a^2-b^{2}P(a\cos \theta ,b\sin \theta )$

$Q(\frac{\pi }{2}+\theta )=\frac{-ax}{\sin \theta }-\frac{by}{\cos \theta }=a^2-b^{2}Q(-a\sin \theta ,b\cos \theta )$

$=\frac{ax}{\sin \theta }+\frac{by}{\cos \theta }=b^2-a^2$

$G(\frac{(a^2-b^2)\cos \theta }{a},0)$

$g(\frac{(b^2-a^2)\sin \theta }{a},0)$

$PG=\sqrt{{(a-\frac{(a^2-b^2)}{a})}^2{\cos }^2\theta +b^2{\sin }^2\theta }$

$=\sqrt{{\left(\frac{b^2}{a}\right)}^2{\cos }^2\theta +b^2{\sin }^2\theta }$

${Qg}^2={(\frac{(b^2-a^2)\sin \theta }{a}+a\sin \theta )}^2+b^2{\cos }^2\theta$

$={\left(\frac{b^2}{a}\sin \theta \right)}^2+b^2{\cos }^2\theta$

${PG}^2+{Qg}^2={\left(\frac{b^2}{a}\right)}^2+b^2$

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

$=a^2(1-e^2)[2-e^2]$