The number of normals that can be drawn throu | vedclass

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

Question

(D) The equation of the parabola is $y^2=4ax$,where $a=1$.Any normal to the parabola $y^2=4ax$ at point $(at^2, 2at)$ has the equation $y = -tx + 2at

Vedclass · Accepted Answer

$3$

Vedclass · Answer

(D) The equation of the parabola is $y^2=4ax$,where $a=1$.Any normal to the parabola $y^2=4ax$ at point $(at^2, 2at)$ has the equation $y = -tx + 2at + at^3$.If this normal passes through $(9,6)$,then $6 = -9t + 2(1)t + (1)t^3$.$6 = -7t + t^3 \Rightarrow t^3 - 7t - 6 = 0$.By testing values,$t=-1$ is a root: $(-1)^3 - 7(-1) - 6 = -1 + 7 - 6 = 0$.Dividing by $(t+1)$,we get $(t+1)(t^2 - t - 6) = 0 \Rightarrow (t+1)(t-3)(t+2) = 0$.The roots are $t = -1, 3, -2$.Since there are $3$ distinct real values for $t$,there are $3$ distinct normals that can be drawn from the point $(9,6)$ to the parabola.

The number of normals that can be drawn through the point $(9,6)$ to the parabola $y^2=4x$ is

Similar Questions

The axis of a parabola is the line $y=x$ and its vertex and focus are in the first quadrant at distances $\sqrt{2}$ and $2\sqrt{2}$ units from the origin,respectively. If the point $(1, k)$ lies on the parabola,then a possible value of $k$ is :-

Find the equation of the parabola with focus $(0, -3)$ and directrix $y = 3$.

If a normal chord of the parabola $y^2 = 4ax$ subtends a right angle at the vertex,its slope is-

The equation of the parabola whose vertex and focus lie on the $x$-axis at distances $a$ and $a'$ from the origin,respectively,is

The locus of the poles of focal chords of the parabola $y^2 = 4ax$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module