If mx - y + c = 0 is a normal at a point P on | vedclass

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

Question

(B) Given,equation of normal $mx - y + c = 0$. Parabola $y^2 = 16x$. Comparing with $y^2 = 4ax$,we get $a = 4$. Let the coordinates of point $P$ be $(

Vedclass · Accepted Answer

$132$

Vedclass · Answer

(B) Given,equation of normal $mx - y + c = 0$. Parabola $y^2 = 16x$. Comparing with $y^2 = 4ax$,we get $a = 4$. Let the coordinates of point $P$ be $(at^2, 2at) = (4t^2, 8t)$. The slope of the tangent at $P$ is $\frac{1}{t}$. The slope of the normal is $m = -t$,so $t = -m$. Thus,the point $P$ is $(4m^2, -8m)$. The focal distance of a point $(at^2, 2at)$ is $a(1 + t^2)$. Given focal distance $= 40$,so $4(1 + t^2) = 40$ $\Rightarrow 1 + t^2 = 10$ $\Rightarrow t^2 = 9$ $\Rightarrow t = \pm 3$. Since $t = -m$,$m = \mp 3$,so $m^2 = 9$. The point $P$ is $(4(9), 8(\mp 3)) = (36, \mp 24)$. Since $P$ lies on the normal $mx - y + c = 0$,we have $m(36) - (\mp 24) + c = 0$. Substituting $m = \mp 3$: $(\mp 3)(36) \pm 24 + c = 0 \Rightarrow \mp 108 \pm 24 + c = 0$. If $m = -3$,then $t = 3$,$P = (36, 24)$,normal is $-3x - y + c = 0$ $\Rightarrow -3(36) - 24 + c = 0$ $\Rightarrow -108 - 24 + c = 0$ $\Rightarrow c = 132$. If $m = 3$,then $t = -3$,$P = (36, -24)$,normal is $3x - y + c = 0$ $\Rightarrow 3(36) - (-24) + c = 0$ $\Rightarrow 108 + 24 + c = 0$ $\Rightarrow c = -132$. In both cases,$|c| = 132$.

If $mx - y + c = 0$ is a normal at a point $P$ on the parabola $y^2 = 16x$ and the focal distance of $P$ is $40$ units,then $|c| =$

Similar Questions

Find the equation of the normal to the curve $x^{2}=4y$ which passes through the point $(1,2)$.

If $(x_1, y_1)$ and $(x_2, y_2)$ are the points on the parabola $y^2 = 32x$ each at a focal distance of $10$ units,then $2(x_1^2 + x_2^2 + y_1^2 + y_2^2) = $

If the tangents drawn at the points $P$ and $Q$ on the parabola $y^{2} = 2x - 3$ intersect at the point $R(0, 1)$,then the orthocentre of the triangle $PQR$ is.

What is the vertex of the parabola $x^2 - 8y - x + 19 = 0$?

If the straight line $y=mx+c$ is parallel to the axis of the parabola $y^2=lx$ and intersects the parabola at $\left(\frac{c^2}{8}, c\right)$,then the length of the latus rectum is

Vedclass Test Series

Exam Paper Generator

Online Exam Module