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BITSAT 2020
Exam
BITSAT
Year
2020
Subject
Mathematics
Chapter
Ellipse

BITSAT 2020 · Mathematics · Ellipse

If x ma + y nb =1 touches the ellipse x^ 2 a^ 2 + y^ 2 b^ 2 =1, then

If xma+ynb=1\frac{\mathrm{x}}{\mathrm{ma}}+\frac{\mathrm{y}}{\mathrm{nb}}=1 touches the ellipse x2a2+y2b2=1,\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1, then

Options

  1. A\mathrmm2=\mathrmn2\mathrmn2−1\mathrmm^{2}=\frac{\mathrmn^{2}}{\mathrmn^{2}-1} or \mathrmn2=\mathrmm2m2−1\mathrmn^{2}=\frac{\mathrmm^{2}}{\mathrm{ m}^{2}-1}
  2. B\mathrmm2=\mathrmn2\mathrmn2+1\mathrmm^{2}=\frac{\mathrmn^{2}}{\mathrmn^{2}+1} or \mathrmn2=\mathrmm2m2+1\mathrmn^{2}=\frac{\mathrmm^{2}}{\mathrm{ m}^{2}+1}
  3. C\mathrmm2=\mathrmn2+1\mathrmn2\mathrmm^{2}=\frac{\mathrmn^{2}+1}{\mathrmn^{2}} or \mathrmn2=\mathrmm2+1m2\mathrmn^{2}=\frac{\mathrmm^{2}+1}{\mathrm{ m}^{2}}
  4. D\mathrmm2=\mathrmn2−1\mathrmn2\mathrmm^{2}=\frac{\mathrmn^{2}-1}{\mathrmn^{2}} or \mathrmn2=\mathrmm2−1m2\mathrmn^{2}=\frac{\mathrmm^{2}-1}{\mathrm{ m}^{2}}

Correct answer

A. \mathrmm2=\mathrmn2\mathrmn2−1\mathrmm^{2}=\frac{\mathrmn^{2}}{\mathrmn^{2}-1} or \mathrmn2=\mathrmm2m2−1\mathrmn^{2}=\frac{\mathrmm^{2}}{\mathrm{ m}^{2}-1}

Step-by-step solution

The line is y=− nb  ma x−nby=-\frac{\text{ nb }}{\text{ ma }} x-n b. It will touch the ellipse if (−nb)2=a2(−nbma)2+b2[c2=a2m2+b2]⇒n2=n2m2+1⇒m2=n2n2−1(-n b)^{2}=a^{2}\left(-\frac{n b}{m a}\right)^{2}+b^{2}\left[\mathrm{c}^{2}=\mathrm{a}^{2} \mathrm{ m}^{2}+\mathrm{b}^{2}\right]\Rightarrow n^{2}=\frac{n^{2}}{m^{2}}+1\Rightarrow \mathrm{m}^{2}=\frac{\mathrm{n}^{2}}{\mathrm{n}^{2}-1} or n2=m2m2−1\mathrm{n}^{2}=\frac{\mathrm{m}^{2}}{\mathrm{ m}^{2}-1}
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