If a normal chord of the parabola y^2 = 4ax s | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) Let the coordinates of the points $P$ and $Q$ on the parabola $y^2 = 4ax$ be $P(at_1^2, 2at_1)$ and $Q(at_2^2, 2at_2)$.Since $PQ$ is a normal chor

Vedclass · Accepted Answer

$\pm \sqrt{2}$

Vedclass · Answer

(B) Let the coordinates of the points $P$ and $Q$ on the parabola $y^2 = 4ax$ be $P(at_1^2, 2at_1)$ and $Q(at_2^2, 2at_2)$.Since $PQ$ is a normal chord,we have the relation $t_2 = -t_1 - \frac{2}{t_1} \quad \dots(1)$.The slopes of $OP$ and $OQ$ are $m_1 = \frac{2at_1}{at_1^2} = \frac{2}{t_1}$ and $m_2 = \frac{2at_2}{at_2^2} = \frac{2}{t_2}$ respectively.Since $PQ$ subtends a right angle at the vertex $O(0,0)$,the product of the slopes of $OP$ and $OQ$ is $-1$.$\frac{2}{t_1} 	imes \frac{2}{t_2} = -1 \Rightarrow t_1t_2 = -4 \quad \dots(2)$.Substituting $t_2$ from $(1)$ into $(2)$,we get $t_1(-t_1 - \frac{2}{t_1}) = -4$.$-t_1^2 - 2 = -4$ $\Rightarrow t_1^2 = 2$ $\Rightarrow t_1 = \pm \sqrt{2}$.The slope of the normal at $P$ is $m = -t_1$.Therefore,the slope of the normal chord is $m = \mp \sqrt{2}$.Since the options represent the set of possible slopes,the correct choice is $\pm \sqrt{2}$.

If a normal chord of the parabola $y^2 = 4ax$ subtends a right angle at the vertex,its slope is-

Similar Questions

The locus of the points of intersection of perpendicular normals to the parabola $y^2=4ax$ is

Let $P : y^{2} = 4ax, a > 0$ be a parabola with focus $S$. Let the tangents to the parabola $P$ that make an angle of $\frac{\pi}{4}$ with the line $y = 3x + 5$ touch the parabola $P$ at $A$ and $B$. Then the value of $a$ for which $A, B$ and $S$ are collinear is:

Find the equation of the parabola that satisfies the following conditions: Focus $(6, 0)$,directrix $x = -6$.

$PQ$ is a focal chord of the parabola $y^2 = 4x$ with focus $S$. If $P = (4, 4)$,then $SQ = $

Let $A(4, -4)$ and $B(9, 6)$ be points on the parabola $y^2 = 4x$. Let $C$ be a point chosen on the arc $AOB$ of the parabola,where $O$ is the origin,such that the area of $\Delta ACB$ is maximum. Then,the area (in sq. units) of $\Delta ACB$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module