Question

If the normal at any given point P on ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ meets its auxiliary circle at Q and R such that $\angle QOR={90}^{\circ }$, where O is centre of ellipse, then

• A. $2(a^2-b^2{)}^2=a^{4}se{c}^2\theta +a^2{b}^{2}cose{c}^2\theta$
• B. $a^4+5a^2{b}^2+2b^4=a^{4}ta{n}^2\theta +a^2{b}^{2}co{t}^2\theta$
• C. $a^4+5b^2{a}^2+2b^4\ge 2a^{3}b$
• D. $a^4+2b^2\ge 5a^2{b}^2+2a^{3}b$

Answer 1

Answer: (A), (D)

The correct options are
A

$2(a^2-b^2{)}^2=a^{4}se{c}^2\theta +a^2{b}^{2}cose{c}^2\theta$

D

$a^4+2b^2\ge 5a^2{b}^2+2a^{3}b$

Homogenizing auxiliary circle with normal at P

$x^2+y^2=\frac{a^2(axsec\theta -\text{by}cosec\theta {)}^2}{(a^2-b^2{)}^2}$

$\angle QOR={90}^{\circ }\Rightarrow -1+\frac{a^{4}se{c}^2\theta }{(a^2-b^2{)}^2}-1+\frac{a^2{b}^{2}cose{c}^2\theta }{(a^2-b^2{)}^2}=0$

$\Rightarrow 2(a^2-b^2{)}^2=a^{4}se{c}^2\theta +a^2{b}^{2}cose{c}^2\theta$

$\Rightarrow 2(a^2-b^2{)}^2=a^4(1+ta{n}^2\theta )+a^2{b}^2(1+co{t}^2\theta )$

$\Rightarrow a^4-5a^2{b}^2+2b^4=a^{4}ta{n}^2\theta +a^2{b}^{2}co{t}^2\theta$

Using AM $\ge$ GM,

$a^{4}ta{n}^2\theta +a^2{b}^{2}co{t}^2\theta =a^{4}ta{n}^2\theta +\frac{a^2{b}^2}{ta{n}^2\theta }\ge 2a^{3}b$