If the line y = -x + k is normal to the curve | vedclass

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

Question

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

Vedclass · Accepted Answer

$12$

Vedclass · Answer

(D) The equation of the parabola is $y^2 = 16x$. Comparing this with $y^2 = 4ax$,we get $a = 4$.The equation of the normal to the parabola $y^2 = 4ax$ with slope $m$ is given by $y = mx - 2am - am^3$.Substituting $a = 4$,the normal is $y = mx - 8m - 4m^3$.Given that the line $y = -x + k$ is a normal,we compare the slopes: $m = -1$.Substituting $m = -1$ into the equation of the normal:$y = (-1)x - 8(-1) - 4(-1)^3$$y = -x + 8 + 4$$y = -x + 12$Comparing this with $y = -x + k$,we get $k = 12$.

If the line $y = -x + k$ is normal to the curve $y^2 = 16x$,then $k$ is

Similar Questions

If the perpendicular distance from the focus of a parabola $y^2=4ax$ to its directrix is $\frac{3}{2}$,then the equation of the normal drawn at $(4a, -4a)$ is

The locus of the middle points of all chords of the parabola $y^2 = 4ax$ passing through the vertex is

The sum of squares of all possible values of $k$,for which the area of the region bounded by the parabolas $2y^2 = kx$ and $ky^2 = 2(y - x)$ is maximum,is equal to:

The shortest distance from $(0,3)$ to the parabola $y^2=4x$ is

The coordinates of a point on the parabola $y^2 = 8x$ whose focal distance is $4$ are:

Vedclass Test Series

Exam Paper Generator

Online Exam Module