The line \(a x+b y=1\) cuts ellipse \(c x^2+d y^2=1\) only once if

The line \(a x+b y=1\) cuts ellipse \(c x^2+d y^2=1\) only once if
  1. \(\mathrm{ca}^2+\mathrm{db}^2=1\)
  2. \(\frac{\mathrm{c}}{\mathrm{a}^2}+\frac{\mathrm{d}}{\mathrm{b}^2}=1\)
  3. \(\frac{\mathrm{a}^2}{\mathrm{c}}+\frac{\mathrm{b}^2}{\mathrm{~d}}=1\)
  4. \(a c^2+b d^2=1\)

Solution

Clearly \(a x+b y=1\) i.e \(y=-\frac{a}{b} x+\frac{1}{b}\) is tangent to $\begin{aligned} & cx^{2}+dy^{2}=1 \Rightarrow \frac{x^{2}}{\frac{1}{c}}+\frac{y^{2}}{\frac{1}{d}}=1 \\ & \therefore\left(\frac{1}{b}\right)^{2}=\left(\frac{1}{c}\right)\left(-\frac{a}{b}\right)^{2}+\left(\frac{1}{d}\right) \\ & \Rightarrow 1=\frac{a^{2}}{c}+\frac{b^{2}}{d} \end{aligned}$

Asked in: BITSAT 2010

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