Mathematics | VITEEE 2015 | Discussion Page

If the line y=2x+c is a normal to the ellipse $\frac{x^{2}}{9}+\frac{y^{2}}{16}=1$ , then

A) $c=\frac{2}{3}$

B) $c=\sqrt{\frac{73}{5}}$

C) $c=\frac{14}{\sqrt{73}}$

D) $c=\sqrt{\frac{5}{7}}$

Option C

if the line y=mx+c isa normal to the ellipse $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$ , then

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

[Here , m=2, a² =9 , and b² =16]

$=\frac{2^{2}(9-16)^{2}}{9+16\times (2)^{2}}$

$=\frac{4\times 49}{9+64}=\frac{4\times 49}{73}=\frac{196}{73}$

$\therefore$ $c=\frac{14}{\sqrt{73}}$

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