Normals are drawn from the point P(8,0) to th | vedclass

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

Question

(B) Given parabola is $y^2=12x$,so $4a=12 \Rightarrow a=3$. Equation of normal with slope $m$ is $y=mx-2am-am^3$,which becomes $y=mx-6m-3m^3$. Since t

Vedclass · Accepted Answer

$2 \sqrt{6}$

Vedclass · Answer

(B) Given parabola is $y^2=12x$,so $4a=12 \Rightarrow a=3$. Equation of normal with slope $m$ is $y=mx-2am-am^3$,which becomes $y=mx-6m-3m^3$. Since the normal passes through $P(8,0)$,we have $0=8m-6m-3m^3$. $2m-3m^3=0 \Rightarrow m(2-3m^2)=0$. The slopes are $m_1=0$,$m_2=\sqrt{\frac{2}{3}}$,and $m_3=-\sqrt{\frac{2}{3}}$. The non-horizontal normals have slopes $m_2=\sqrt{\frac{2}{3}}$ and $m_3=-\sqrt{\frac{2}{3}}$. The angle $	heta$ between these two normals is given by $	an 	heta = \left| \frac{m_2-m_3}{1+m_2m_3} ight|$. $	an 	heta = \left| \frac{\sqrt{\frac{2}{3}} - (-\sqrt{\frac{2}{3}})}{1 + (\sqrt{\frac{2}{3}})(-\sqrt{\frac{2}{3}})} ight| = \left| \frac{2\sqrt{\frac{2}{3}}}{1 - \frac{2}{3}} ight| = \frac{2\sqrt{\frac{2}{3}}}{\frac{1}{3}} = 6 \sqrt{\frac{2}{3}} = 6 \frac{\sqrt{2}}{\sqrt{3}} = 2\sqrt{6}$.

Normals are drawn from the point $P(8,0)$ to the parabola $y^2=12x$. If $\theta$ is the acute angle between two non-horizontal normals among them,then $\tan \theta=$

Similar Questions

$A$ tangent is drawn to the parabola $y^{2}=6x$ which is perpendicular to the line $2x+y=1$. Which of the following points does $NOT$ lie on it?

If $m_1$ and $m_2$ are the slopes of the tangents drawn from the point $(1, 4)$ to the parabola $y^2 = 11x$,then $2(m_1^2 + m_2^2) = $

Find the coordinates of the focus,axis of the parabola,the equation of the directrix,and the length of the latus rectum for $y^{2}=10x$.

The equation of a tangent to the parabola,$x^2 = 8y,$ which makes an angle $\theta$ with the positive direction of the $x-$axis,is

Tangents are drawn at three points $P(t_1), Q(t_2), R(t_3)$ on the parabola $y^2 = x$. Let these tangents intersect each other at the points $L, M, N$. If $t_1 = 2, t_2 = -4, t_3 = 6$,then the area of the triangle $LMN$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module