- A$ - \alpha , - \beta $
- B$\alpha ,\frac{1}{\beta }$
- ✓$\frac{1}{\alpha },\frac{1}{\beta }$
- DNone of these
==> $\alpha + \beta = - \frac{b}{a}$ and $\alpha \beta = \frac{c}{a}$
Let the roots of $c{x^2} + bx + a = 0$be $\alpha ',\beta '$, then
$\alpha ' + \beta ' = - \frac{b}{c}$and $\alpha '\beta ' = \frac{a}{c}$
but $\frac{{\alpha + \beta }}{{\alpha \beta }} = \frac{{ - b/a}}{{c/a}} = \frac{{ - b}}{c}$
==>$\frac{1}{\alpha } + \frac{1}{\beta } = \alpha ' + \beta '$
Hence $\alpha ' = \frac{1}{\alpha }$ and $\beta ' = \frac{1}{\beta }$.
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$L _1: x \sqrt{2}+ y -1=0$ and $L _2: x \sqrt{2}- y +1=0$
For a fixed constant $\lambda$, let $C$ be the locus of a point $P$ such that the product of the distance of $P$ from $L_1$ and the distance of $P$ from $L_2$ is $\lambda^2$. The line $y=2 x+1$ meets $C$ at two points $R$ and $S$, where the distance between $R$ and $S$ is $\sqrt{270}$.
Let the perpendicular bisector of $RS$ meet $C$ at two distinct points $R ^{\prime}$ and $S ^{\prime}$. Let $D$ be the square of the distance between $R ^{\prime}$ and $S ^{\prime}$.
($1$) The value of $\lambda^2$ is
($2$) The value of $D$ is