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

(B) The equation of the parabola is $y^2 = 16x$. Comparing this with $y^2 = 4ax$,we get $4a = 16$,so $a = 4$.The equation of a normal to the parabola

Vedclass · Accepted Answer

$4x + 16y = 33$

Vedclass · Answer

(B) The equation of the parabola is $y^2 = 16x$. Comparing this with $y^2 = 4ax$,we get $4a = 16$,so $a = 4$.The equation of a normal to the parabola $y^2 = 4ax$ with slope $m$ is given by $y = mx - 2am - am^3$.Given $m = -1/4$ and $a = 4$,we substitute these values into the formula:$y = (-1/4)x - 2(4)(-1/4) - 4(-1/4)^3$$y = -x/4 + 2 - 4(-1/64)$$y = -x/4 + 2 + 1/16$Multiply the entire equation by $16$ to clear the denominators:$16y = -4x + 32 + 1$$16y = -4x + 33$$4x + 16y = 33$

Find the equation of the normal to the parabola $y^2 = 16x$ having a slope of $-1/4$.

Similar Questions

The length of the latus rectum of a parabola whose directrix is $x + y - 2 = 0$ and focus is $(3, -4)$ is:

The normal to the curve $x^{2}=4y$ passing through $(1,2)$ is

If a normal chord at a point $t$ on the parabola $y^2=4ax$ subtends a right angle at the vertex,then $t^2$ equals to

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) = $

$A$ ray of light moving parallel to the $x$-axis gets reflected from a parabolic mirror whose equation is $(y - 2)^2 = 4(x + 1)$. After reflection,the ray must pass through the point:

Vedclass Test Series

Exam Paper Generator

Online Exam Module