Question

A focal chord of parabola $y^2=4x$ is inclined at an angle of $\frac{\pi }{4}$ with the positive direction of x - axis, then the slope of normal drawn at the ends of focal chord will satisfy the equation :

• A. $m^2-2m-1=0$
• B. $m^2+2m-1=0$
• C. $m^2-1=0$
• D. none of these

Answer 1

Answer: (B)

$m^2+2m-1=0$

Let $A,B$ be the points $(t_1^2,2t_1)$ and $(t_2^2,2t_2),(a=1)$ be two points on the parabola $y^2=4x.$
Since $AB$ is a focal chord, $t_1{t}_2=-1.$
Also slope of chord $y(t_1+t_2)-2x-2a{t}_1{t}_2=0$ is
$\tan \frac{\pi }{4}=1=\frac{2}{t_1+t_2}\therefore t_1+t_2=2$
Hence $t_1,t_2$ are the roots of
$m^2-2m-1=0$ ...(1)
Slopes of normals at $A$ and $B$ are $-t_1,-t_2$ which are roots of ${(-m)}^2-2(-m)-1=0$
or $m^2+2m-1=0$