Question

The line $ax+by=1$ cuts ellipse $c{x}^2+d{y}^2=1$ only once if

• A. $c{a}^2+d{b}^2=1$
• B. $\frac{c}{a^2}+\frac{d}{b^2}=1$
• C. $\frac{a^2}{c}+\frac{b^2}{d}=1$
• D. $a{c}^2+b{d}^2=1$

Answer 1

Answer: (C)

Clearly $ax+by=1$
i.e $y=-\frac{a}{b}x+\frac{1}{b}$ is tangent to
$c{x}^2+d{y}^2=1$
$\Rightarrow \frac{x^2}{\frac{1}{c}}+\frac{y^2}{\frac{1}{d}}=1$
$\therefore {\left(\frac{1}{b}\right)}^2=\left(\frac{1}{c}\right){\left(-\frac{a}{b}\right)}^2+\left(\frac{1}{d}\right)$
$\Rightarrow 1=\frac{a^2}{c}+\frac{b^2}{d}$