If the normal drawn at P(8, 16) to the parabo | vedclass

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

Question

(A) The equation of the parabola is $y^2 = 32x$,so $4a = 32$,which implies $a = 8$. Point $P$ is $(8, 16)$,which corresponds to $(at^2, 2at) = (8t^2,

Vedclass · Accepted Answer

$x + 3y + 72 = 0$

Vedclass · Answer

(A) The equation of the parabola is $y^2 = 32x$,so $4a = 32$,which implies $a = 8$. Point $P$ is $(8, 16)$,which corresponds to $(at^2, 2at) = (8t^2, 16t)$. Thus,$t_1 = 2$. The normal at $t_1$ meets the parabola at $t_2 = -t_1 - \frac{2}{t_1} = -2 - \frac{2}{2} = -3$. The coordinates of $Q$ are $(at_2^2, 2at_2) = (8(-3)^2, 16(-3)) = (72, -48)$. The equation of the tangent at $Q(x_1, y_1)$ to $y^2 = 4ax$ is $yy_1 = 2a(x + x_1)$. Substituting $y_1 = -48$,$a = 8$,and $x_1 = 72$: $y(-48) = 2(8)(x + 72)$ $-48y = 16(x + 72)$ $-3y = x + 72$ $x + 3y + 72 = 0$.

If the normal drawn at $P(8, 16)$ to the parabola $y^2 = 32x$ meets the parabola again at $Q$,then the equation of the tangent drawn at $Q$ to the parabola is

Similar Questions

The normal at a point on the parabola $y^2=4x$ passes through $(5,0)$. If there are two more normals to this parabola passing through $(5,0)$,then the equation of one of these normals is

Let $P, Q,$ and $R$ be three co-normal points on the parabola $y^2 = 4ax$. Then the correct statement$(s)$ is/are:

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

The equation of a tangent to the parabola $y^2 = 8x$ is $y = x + 2$. The point on this line from which the other tangent to the parabola is perpendicular to the given tangent is

The axis of a parabola lies along the $x$-axis. If its vertex and focus are at distances $2$ and $4$ respectively from the origin on the positive $x$-axis,then which of the following points does not lie on it?

Vedclass Test Series

Exam Paper Generator

Online Exam Module