1. Tardigrade
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  3. Mathematics
  4. The circle x2+y2=4 cuts the line joining the points A(1 , 0) and B(3 , 4) in two points P and Q . Let, (P B/P A)=α and (B Q/A Q)=β , then which of the following quadratic equation has either α or β as one of their root

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Q. The circle $x^{2}+y^{2}=4$ cuts the line joining the points $A\left(1 , 0\right)$ and $B\left(3 , 4\right)$ in two points $P$ and $Q$ . Let, $\frac{P B}{P A}=\alpha $ and $\frac{B Q}{A Q}=\beta $ , then which of the following quadratic equation has either $\alpha $ or $\beta $ as one of their root

NTA AbhyasNTA Abhyas 2022

Solution:

Solution
$\because \frac{B P}{P A}=\alpha $
$BP:PA=\alpha :1$
$\therefore $ Coordinates of $P$ is $\left(\frac{3 + \alpha }{1 + \alpha } , \frac{4}{\alpha + 1}\right)$
$P$ lie on $x^{2}+y^{2}=4$
$\Rightarrow(\alpha+3)^2+16=4(\alpha+1)^2$
$\Rightarrow 3\left(\alpha \right)^{2}+2\alpha -21=0\ldots \left(i\right)$
and $\frac{B Q}{Q A}=\frac{\beta }{1}$
$\Rightarrow BQ:QA=\beta :1$
$\Rightarrow \frac{B Q}{Q A}-1=\beta -1$
$\Rightarrow \frac{A B}{Q A}=\frac{\left(\right. \beta - 1 \left.\right)}{1}$
$\Rightarrow AB:QA=\left(\right.\beta -1\left.\right):1$
$\therefore $ Coordinates of $Q$ is $\left(\frac{\beta - 3}{\beta - 1} , - \frac{4}{\beta - 1}\right)$
$Q$ lie on $x^{2}+y^{2}=4$
$\therefore(\beta-3)^2+16=4(\beta-1)^2$
$\Rightarrow 3\beta ^{2}-21\beta -21=0$
Hence, $\alpha $ is a roots of $3x^{2}+2x-21=0$ and
$\beta $ is a root of $3x^{2}-2x-21=0$ .