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 area of the triangle formed by the lines joining the vertex of the parabola $x^2 = 12y$ to the ends of its latus rectum is .................. $sq. \ unit$.

Find the equation of the directrix of the parabola $y^2 = 12x$.

The largest value of $k$ for which the circle $x^2+y^2=k^2$ lies completely in the interior of the parabola $y^2=4x+16$ is

If the normal chord drawn at the point $\left(\frac{15}{2}, \frac{15}{\sqrt{2}}\right)$ to the parabola $y^2=15x$ subtends an angle $\theta$ at the vertex of the parabola,then $\sin \frac{\theta}{3}+\cos \frac{2\theta}{3}-\sec \frac{4\theta}{3}=$

Three normals to the parabola $y^2 = x$ are drawn through a point $(C, 0)$. Then:

Vedclass Test Series

Exam Paper Generator

Online Exam Module