MCQ
Two parabolas have the same focus $(4,3)$ and their directrices are the $x$-axis and the $y$-axis, respectively. If these parabolas intersects at the points $A$ and $B$, then $(A B)^{2}$ is equal to
- ✓192
- B384
- C96
- D392
✓
Answer
Correct option: A.
192
(A)
Let intersection points of these two parabolas are
$\mathrm{A}\left(\mathrm{x}_{1}, \mathrm{y}_{1}\right) \& \mathrm{~B}\left(\mathrm{x}_{2}, \mathrm{y}_{2}\right)$
$\because$ equation of parabola I and II are given below
$\therefore(\mathrm{x}-4)^{2}+(\mathrm{y}-3)^{2}=\mathrm{x}^{2}$
$\&(x-4)^{2}+(y-3)^{2}=y^{2}$
Here $A\left(x_{1}, y_{1}\right) \& B\left(x_{2}, y_{2}\right)$ will satisfy the equation
Also from equations (1) \& (2), we get $x=y$..(3)
Put $\mathrm{x}=\mathrm{y}$ in equation (1)
We get $x^{2}-14 x+25=0$
$\mathrm{x}_{1}+\mathrm{x}_{2}=14$
$\mathrm{x}_{1} \mathrm{X}_{2}=25$
$\therefore \mathrm{AB}^{2}=\left(\mathrm{x}_{1}-\mathrm{x}_{2}\right)^{2}+\left(\mathrm{y}_{1}-\mathrm{y}_{2}\right)^{2}$
$=2\left(x_{1}-x_{2}\right)^{2}$
$=2\left[\left(x_{1}+x_{2}\right)^{2}-4 x_{1} x_{2}\right]$
$=192$
Let intersection points of these two parabolas are
$\mathrm{A}\left(\mathrm{x}_{1}, \mathrm{y}_{1}\right) \& \mathrm{~B}\left(\mathrm{x}_{2}, \mathrm{y}_{2}\right)$
$\because$ equation of parabola I and II are given below
$\therefore(\mathrm{x}-4)^{2}+(\mathrm{y}-3)^{2}=\mathrm{x}^{2}$
$\&(x-4)^{2}+(y-3)^{2}=y^{2}$
Here $A\left(x_{1}, y_{1}\right) \& B\left(x_{2}, y_{2}\right)$ will satisfy the equation
Also from equations (1) \& (2), we get $x=y$..(3)
Put $\mathrm{x}=\mathrm{y}$ in equation (1)
We get $x^{2}-14 x+25=0$
$\mathrm{x}_{1}+\mathrm{x}_{2}=14$
$\mathrm{x}_{1} \mathrm{X}_{2}=25$
$\therefore \mathrm{AB}^{2}=\left(\mathrm{x}_{1}-\mathrm{x}_{2}\right)^{2}+\left(\mathrm{y}_{1}-\mathrm{y}_{2}\right)^{2}$
$=2\left(x_{1}-x_{2}\right)^{2}$
$=2\left[\left(x_{1}+x_{2}\right)^{2}-4 x_{1} x_{2}\right]$
$=192$
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