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BITSAT 2010
Exam
BITSAT
Year
2010
Subject
Mathematics
Chapter
Circle

BITSAT 2010 · Mathematics · Circle

The line a x+b y=1 cuts ellipse c x²+d y²=1 only once if

The line ax+by=1a x+b y=1 cuts ellipse cx2+dy2=1c x²+d y²=1 only once if

Options

  1. A(\mathrm{ca}^2+\mathrm{db}^2=1)
  2. B(ca2+db2=\frac{\mathrm{c}}{\mathrm{a}^2}+\frac{\mathrm{d}}{\mathrm{b}^2}=1)
  3. C(a2c+b2 d=\frac{\mathrm{a}^2}{\mathrm{c}}+\frac{\mathrm{b}^2}{\mathrm{~d}}=1)
  4. Dac2+bd2=1a c²+b d²=1

Correct answer

C. (a2c+b2 d=\frac{\mathrm{a}^2}{\mathrm{c}}+\frac{\mathrm{b}^2}{\mathrm{~d}}=1)

Step-by-step solution

Clearly ax+by=1)i.e(y=−abx+1b)istangenta x+b y=1) i.e (y=-\frac{a}{b} x+\frac{1}{b}) is tangent to cx2+dy2=1⇒x21c+y21d=1∴(1b)2=(1c)(−ab)2+(1d)⇒1=a2c+b2d\begin{aligned} \mathrm{cx}^2+\mathrm{dy}^2=1 \Rightarrow \frac{\mathrm{x}^2}{\frac{1}{c}}+\frac{\mathrm{y}^2}{\frac{1}{d}}=1 \\ & \therefore\left(\frac{1}{b}\right)^2=\left(\frac{1}{c}\right)\left(-\frac{a}{b}\right)^2+\left(\frac{1}{d}\right)\\ & \Rightarrow 1=\frac{a^2}{c}+\frac{b^2}{d} \end{aligned}
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