Question

The tangent and normal to the parabola $y^2=8x$ drawn at $(2,4)$ intersect the line $ix+y=3$ at the points A and B, respectively. If AB subtends a right angle at the origin, then the sum of all possible values of $i$ is

• A. $2$
• B. $1$
• C. $0$
• D. $-1$

Answer 1

Answer: (D)

$-1$

Given Parabola: $y^2=8x.........\left(i\right)$

Tangent and normal drawn drawn at $P(2,4)$ intersect the line.

$=>L_1:+y=ix+y=3.......\left(ii\right)$ at point $A$ and $B$

Tangent at $(2,4)=>4y=\frac{8(x+2)}{2}$

$T=>y=x+2........\left(iii\right)$

Equation of norml at $(2,4)=>N:(y-4)=\frac{-4}{4}(x-2)$

$=>N:y=-x+6........\left(iv\right)$

Point of intersection of $T$ and $L_1$ $=>$ $A:(\frac{1}{i+1},\frac{3+2i}{i+1})$

Point of intersection of $N$ and $L_1$ $=>$$B:(\frac{3}{1-i},\frac{3-6i}{1-i})$

$AB$ sub tends right angle at $(0,0)$.

Line through $(0,0)$ and $A$ $=>$ $y=\left(\frac{\frac{3+2i}{i+1}}{\frac{1}{i+1}}\right)(x-0)$

$=>y=(1-2i)x$

So, $m_1=3+2i$ and $m_2=1-2i$

So, tangent of angle between them, $tan\left(\frac{\pi }{2}\right)=\frac{m_1-m_2}{1-m_1{m}_2}$

$=>\frac{1}{0}=\frac{m_1-m_2}{1-m_1{m}_2}$

$=>1-m_1{m}_2=0$

$=>1=(1-2i)(3-2i)$

$=>1=3-4i-4i^2$

$=>4i^2+4i-2=0($quadratic$)$

Sum of possible values of $i=$ Sum of roots of above equation$=\frac{-4}{4}=-1$.