If the normal chord drawn at (2 a, 2 a sqrt{2 | vedclass

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

Question

(B) The parabola is $y^2 = 4ax$. The point $P$ is $(2a, 2a\sqrt{2})$.Comparing $P(2a, 2a\sqrt{2})$ with $(at^2, 2at)$,we get $t = \sqrt{2}$.The normal

Vedclass · Accepted Answer

$90$

Vedclass · Answer

(B) The parabola is $y^2 = 4ax$. The point $P$ is $(2a, 2a\sqrt{2})$.Comparing $P(2a, 2a\sqrt{2})$ with $(at^2, 2at)$,we get $t = \sqrt{2}$.The normal at $t$ meets the parabola at $t_1 = -t - \frac{2}{t} = -\sqrt{2} - \frac{2}{\sqrt{2}} = -2\sqrt{2}$.The coordinates of $Q$ are $(at_1^2, 2at_1) = (a(-2\sqrt{2})^2, 2a(-2\sqrt{2})) = (8a, -4a\sqrt{2})$.The vertex is $O(0, 0)$.The slope of $OP$ is $m_1 = \frac{2a\sqrt{2}}{2a} = \sqrt{2}$.The slope of $OQ$ is $m_2 = \frac{-4a\sqrt{2}}{8a} = -\frac{\sqrt{2}}{2} = -\frac{1}{\sqrt{2}}$.Since $m_1 	imes m_2 = \sqrt{2} 	imes (-\frac{1}{\sqrt{2}}) = -1$,the lines $OP$ and $OQ$ are perpendicular.Therefore,$	heta = 90^{\circ}$.

If the normal chord drawn at $(2 a, 2 a \sqrt{2})$ on the parabola $y^2=4 a x$ subtends an angle $\theta$ at its vertex,then $\theta=$ (in $^{\circ}$)

Similar Questions

If the line $x + y - 1 = 0$ is tangent to the parabola $y^2 = kx$,find the value of $k$.

If $L$ is the normal drawn to the parabola $y^2 = 8x$ at the point $t = \frac{1}{\sqrt{2}}$,then the foot of the perpendicular drawn from the focus of the parabola on to the normal $L$ is

At what point on the parabola $y^2 = 4x$ does the normal make equal angles with the coordinate axes?

The equations of the tangents to the parabola $y^2 = 4ax$ at the ends of its latus rectum are-

The equation of the latus rectum of a parabola is $x+y=8$ and the equation of the tangent at the vertex is $x+y=12$. Then the length of the latus rectum is

Vedclass Test Series

Exam Paper Generator

Online Exam Module