Q. If $a \neq 0$ and the line $2bx + 3cy + 4d = 0,$ passes through the points of intersection of the parabolas$y^{2} = 4ax\ x^{2} = 4ay$, then
A) $d^{2} + (2b + 3c)^{2} = 0$
B) $d^{2} + (3b + 2c)^{2} = 0$
C) $d^{2} + (2b - 3c)^{2} = 0$
D) $d^{2} + (3b - 2c)^{2} = 0$
Explanation
(a)
$y - 0 = \frac{\frac{8}{5}}{- \frac{9}{5} - 3}(x - 3) = \frac{8}{- 24}(x - 3)$
[image:11_image57.emf]
$\Rightarrow y = \ - \frac{1}{3}(x - 3)$
$\Rightarrow \ x + 3y = 3\$ …(i)
Equation of the straight line perpendicular to $AB$ through
$B\ \left( \ - \frac{9}{5},\frac{8}{5} \right)$ is$\ y = \frac{8}{5}$
Hence, the orthocentre is $\left( \frac{11}{5},\frac{8}{5} \right)$
(iii) Equation of $AB$ is $y - 0 = - \frac{1}{3}(x - 3)$
$x + 3y - 3 = 0$
$|x + 3y - 3|^{2} = 10\ \lbrack(x - 3)^{2} + \left( y - {4)}^{2} \right\rbrack$
(Look at coefficient of $x^{2}$ and $y^{2}$ in the answers)
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.