Find the equation of the normal to the parabo | vedclass

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

Question

(D) Step $1$: Find the upper end of the latus rectum for the given parabola.The coordinates of the ends of the latus rectum of the parabola $y^2 = -4a

Vedclass · Accepted Answer

$x - y + 9 = 0$

Vedclass · Answer

(D) Step $1$: Find the upper end of the latus rectum for the given parabola.The coordinates of the ends of the latus rectum of the parabola $y^2 = -4ax$ are $(-a, \pm 2a)$.Comparing with the given parabola $y^2 = -12x$,we get $4a = 12$,so $a = 3$.Hence,the upper end of its latus rectum is $(-3, 6)$.Step $2$: Find the slope of the normal to the given parabola at this point.Given equation of the parabola is $y^2 = -12x$.Differentiating both sides with respect to $x$,we get $2y \frac{dy}{dx} = -12$,which implies $\frac{dy}{dx} = \frac{-6}{y}$.The slope of the tangent at $(-3, 6)$ is $m_t = \frac{-6}{6} = -1$.The slope of the normal at $(-3, 6)$ is $m_n = -\frac{1}{m_t} = -\frac{1}{-1} = 1$.Step $3$: Find the equation of the normal.The equation of the normal is a line passing through $(-3, 6)$ with slope $m = 1$.Using the point-slope form $y - y_1 = m(x - x_1)$:$y - 6 = 1(x - (-3))$$y - 6 = x + 3$$x - y + 9 = 0$.

Find the equation of the normal to the parabola $y^2 + 12x = 0$ at the upper end of its latus rectum.

Similar Questions

If tangents are drawn from the point $(-1, 2)$ to the parabola $y^2 = 4x$, what is the area of the triangle formed by the chord of contact and the tangents (in $\sqrt{2}$)?

Tangents are drawn from the point $(-1, 2)$ to the parabola $y^2 = 4x$. The length of the intercept made by these tangents on the line $x = 2$ is:

The equation of the parabola whose vertex is $(-1, -2)$,axis is vertical and which passes through the point $(3, 6)$,is

If the point $(a, 2a)$ is an interior point of the region bounded by the parabola $y^2 = 16x$ and the double ordinate through the focus,then

If $x+5=0$ is the directrix and $(-3,0)$ is the vertex of a parabola,then the equation of this parabola is . . . . . .

Vedclass Test Series

Exam Paper Generator

Online Exam Module