JEE Main 2017 · 9 Apr · Shift 1 · Mathematics Q81

← JEE Main 2017 · 9 Apr · Shift 1

MathematicsSingle correct+4 / -1Question 81 of 90

If $y=mx+c$ is the normal at a point on the parabola $y^2=8x$ whose focal distance is 8 units, then |c| is equal to :

2\sqrt3

8\sqrt3

10\sqrt3

16\sqrt3

Answer and worked solution

Official answer: C

For $y^2=8x$ , $a=2$ . A focal distance of 8 gives $x+2=8$ , so the point is $(6,\pm4\sqrt3)$ and $t=\pm\sqrt3$ . The normal $y=-tx+2at+at^3$ has
$|c|=4\sqrt3+6\sqrt3=10\sqrt3.$

Topics:Parabola Tangents and normals to conics

← Previous question Next question →

JEE Main 2017 · 9 Apr · Shift 1, Mathematics question 81