Question

Maximum length of perpendicular from centre of ellipse $\frac{x^2}{9}+\frac{y^2}{4}=1$ on any normal to this ellipse is equal to $a+5$, then value of $a$ is

• A. $-2$
• B. $-5$
• C. $-4$
• D. None of the above

Answer 1

The correct option is C $-4$

$\frac{x^2}{9}+\frac{y^2}{4}=1---\left(i\right)$

Normal at any point on $\left(i\right)$ in parametric form is$ax\sec \varphi -bycosec\varphi =a^2-b^2$

$\Rightarrow \left(3\sec \varphi \right)x-\left(2cosec\varphi \right)y=5---\left(ii\right)$

now, perpendicular from $(0,0)$ on $ii$ $\Rightarrow A=\frac{\left|3\sec \varphi \right(0)-(2cosec\varphi \left)\right(0)-5|}{\sqrt{9\sec {}^2\varphi +4cose{c}^2\varphi }}$

for $A$=maximum, $\frac{dA}{d\varphi }=0$

$\Rightarrow \frac{d}{d\varphi }\left(\frac{5}{\sqrt{9\sec {}^2\varphi +4cose{c}^2\varphi }}\right)=0$

$\Rightarrow -\frac{1}{2}\frac{5(18\sec \varphi .\sec \varphi \tan \varphi +8cosec\varphi (-cosec\varphi \cot \varphi ))}{(9\sec {}^2\varphi +4cose{c}^2\varphi {)}^{\frac{3}{2}}}=0$

$\Rightarrow \frac{18sin\varphi }{cos{}^2\varphi cos\varphi }-\frac{8}{si{n}^2\varphi }.\frac{cos\varphi }{sin\varphi }=0$

$\Rightarrow \frac{18si{n}^4\varphi -8cos{}^4\varphi }{{\cos }^4\varphi {\sin }^3\varphi }=0$

$\Rightarrow A=\frac{|+5|}{\sqrt{9{\left(1\right)}^2+4(2{)}^2}}$

$\Rightarrow A=\frac{5}{5}$

$\Rightarrow a+5=1$

$\Rightarrow a=-4$