Q48 / 120 · VITEEE 2015 · Mathematics

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

  • A
    $c=\frac{2}{3}$
  • B
    $c=\sqrt{\frac{73}{5}}$
  • Correct answer
    C
    $\quad c=\frac{14}{\sqrt{73}}$
  • D
    $c=\sqrt{\frac{5}{7}}$

Solution

If the line $y=m x+c$ is a normal to the ellipse $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$, then $\mathrm{c}^{2}=\frac{\mathrm{m}^{2}\left(\mathrm{a}^{2}-\mathrm{b}^{2}\right)^{2}}{\mathrm{a}^{2}+\mathrm{b}^{2} \mathrm{~m}^{2}}$ $\left[\right.$ Here, $\mathrm{m}=2, \mathrm{a}^{2}=9$ and $\left.\mathrm{b}^{2}=16\right]$ $=\frac{(2)^{2}(9-16)^{2}}{9+16 \times(2)^{2}}$ $=\frac{4 \times 49}{9+64}=\frac{4 \times 49}{73}=\frac{196}{73}$ $\therefore \quad c=\frac{14}{\sqrt{73}}$

Hence, the correct answer is option: 'C'.

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