ParabolaHardBloom L3
Question
$A$ is a point on the parabola $y^{2} = 4ax$. The normal at $A$ cuts the parabola again at point $B$. If $AB$ subtends a right angle at the vertex of the parabola, then the slope of $AB$ is :
Options
A.2
B.$\sqrt{2}$
C.$- \sqrt{2}$
D.None of these
Solution
Sol. [Key Concept: The normal at $A(t)$ meets again at $B(t')$ with $t' = - t - \frac{2}{t}$; $AB$ subtending a right angle at the vertex gives $t\ t' = - 4$, fixing the slope $\frac{2}{t + t'}$.]
$$\frac{2}{t_{1}}\frac{2}{t_{2}} = - 1 \Rightarrow t_{1}t_{2} = - 4$$
$$\begin{matrix} & t_{2} = - t_{1} - \frac{2}{t_{1}} \Rightarrow \frac{- 4}{t_{1}} = - t_{1} - \frac{2}{t_{1}} \\ & t_{1} = \pm \sqrt{2} \end{matrix}$$
∴ slope of $AB = - t_{1} = \mp \sqrt{2}$
Create a free account to view solution
View Solution FreeMore Parabola Questions
If x + y = k is normal to y2 =12x , then k is...Let the point P be the vertex of the parabola $y = x^{2} - 6x + 12$. If a line passing through the point $P$ intersects ...If the chord joining the points $P_1(x_1, y_1)$ and $P_2(x_2, y_2)$ on the parabola $y^2 = 12x$ subtends a right angle a...A point on the parabola y2 = 18x at which the ordinate increases at twice the rate of the abscissa is...A parabola $y = ax^{2} + bx + c$ crosses the $x$-axis at $(\alpha,0)(\beta,0)$ both to the right of the origin. A circle...