Q53 / 125 · BITSAT 2010 · Mathematics
The line \(a x+b y=1\) cuts ellipse \(c x^2+d y^2=1\) only once if
- A\(\mathrm{ca}^2+\mathrm{db}^2=1\)
- B\(\frac{\mathrm{c}}{\mathrm{a}^2}+\frac{\mathrm{d}}{\mathrm{b}^2}=1\)
- Correct answerC\(\frac{\mathrm{a}^2}{\mathrm{c}}+\frac{\mathrm{b}^2}{\mathrm{~d}}=1\)
- D\(a c^2+b d^2=1\)
Solution
Clearly \(a x+b y=1\)
i.e \(y=-\frac{a}{b} x+\frac{1}{b}\) is tangent to
\(\begin{aligned}
& \mathrm{cx}^2+\mathrm{dy}^2=1 \Rightarrow \frac{\mathrm{x}^2}{\frac{1}{c}}+\frac{\mathrm{y}^2}{\frac{1}{d}}=1 \\
& \therefore\left(\frac{1}{b}\right)^2=\left(\frac{1}{c}\right)\left(-\frac{a}{b}\right)^2+\left(\frac{1}{d}\right) \\
& \Rightarrow 1=\frac{a^2}{c}+\frac{b^2}{d}
\end{aligned}\)
Hence, the correct answer is option: 'C'.