The normal at the point $(bt_1^2, 2bt_1)$ on the parabola $y^2 = 4bx$ meets the parabola again at the point $(bt_2^2, 2bt_2)$. Then:

Let $A(1, 2)$,$B(4, -4)$,and $C(2, 2\sqrt{2})$ be points on the parabola $y^2 = 4x$. If $\alpha$ and $\beta$ respectively represent the area of $\triangle ABC$ and the area of the triangle formed by the tangents at $A, B, C$ to the parabola,then $\alpha \beta =$

The equation of a tangent to the parabola $y^2 = 8x$ is $y = x + 2$. The point on this line from which the other tangent to the parabola is perpendicular to the given tangent is

Let $L$ be a normal to the parabola $y^2=4x$. If $L$ passes through the point $(9,6)$,then $L$ is given by
$(A)$ $y-x+3=0$ $(B)$ $y+3x-33=0$ $(C)$ $y+x-15=0$ $(D)$ $y-2x+12=0$

$A$ circle of radius $4$,drawn on a chord of the parabola $y^2 = 8x$ as diameter,touches the axis of the parabola. Then,the slope of the chord is

The equation of the directrix of the parabola ${x^2} + 8y - 2x = 7$ is