Question

Let $L _{1}$ be a tangent to the parabola $y ^{2}=4( x +1)$ and $L _{2}$ be a tangent to the parabola $y ^{2}=8( x +2)$ such that $L _{1}$ and $L _{2}$ intersect at right angles. Then $L _{1}$ and $L _{2}$ meet on the straight line:-

• A. $x+3=0$
• B. $x+2 y=0$
• C. $2 x +1=0$
• D. $x+2=0$

Answer 1

Answer: (A)

$y^{2}=4(x+1)$
equation of tangent $y=m(x+1)+\frac{1}{m}$
$y = mx + m +\frac{1}{ m }$
$y^{2}=8(x+2)$
equation of tangent $y = m '( x +2)+\frac{2}{ m' }$
$y = m' x +2\left( m '+\frac{1}{ m '}\right)$
since lines intersect at right angles
$\therefore mm'=-1$
Now $y=m x+m+\frac{1}{m} ....(1)$
$y = m' x +2\left( m'+\frac{1}{ m '}\right)$
$y =-\frac{1}{ m } x +2\left(-\frac{1}{ m }- m \right)$
$y =-\frac{1}{ m } x -2\left( m +\frac{1}{ m }\right) .... (2)$
From equation (1) and (2)
$mx + m +\frac{1}{ m }=-\frac{1}{ m } x -2\left( m +\frac{1}{ m }\right)$
$\left(m+\frac{1}{m}\right) x+3\left(m+\frac{1}{m}\right)=0$
$\therefore x+3=0$