Let P be the point (2, 0) and Q be a variable | vedclass

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

Question

(D) Let the coordinates of point $Q$ on the parabola $(y - 6)^2 = 2(x - 4)$ be $(4 + \frac{t^2}{2}, 6 + t)$.Given $P = (2, 0)$.Let the mid-point of $P

Vedclass · Accepted Answer

$y^2 - x - 6y + 12 = 0$

Vedclass · Answer

(D) Let the coordinates of point $Q$ on the parabola $(y - 6)^2 = 2(x - 4)$ be $(4 + \frac{t^2}{2}, 6 + t)$.Given $P = (2, 0)$.Let the mid-point of $PQ$ be $R(h, k)$.Then $h = \frac{2 + 4 + \frac{t^2}{2}}{2} = 3 + \frac{t^2}{4}$ and $k = \frac{0 + 6 + t}{2} = 3 + \frac{t}{2}$.From the second equation,$\frac{t}{2} = k - 3$,so $t = 2(k - 3)$.Substituting $t$ into the equation for $h$:$h = 3 + \frac{(2(k - 3))^2}{4} = 3 + \frac{4(k - 3)^2}{4} = 3 + (k - 3)^2$.$h - 3 = (k - 3)^2$.Replacing $(h, k)$ with $(x, y)$,we get $(y - 3)^2 = x - 3$.$y^2 - 6y + 9 = x - 3$.$y^2 - 6y - x + 12 = 0$.

Let $P$ be the point $(2, 0)$ and $Q$ be a variable point on the parabola $(y - 6)^2 = 2(x - 4)$. Then the locus of the mid-point of $PQ$ is:

Similar Questions

Find the equation of the chord of the parabola $y^2 = 6x$ passing through the vertex and the negative end of the latus rectum.

If the point on the curve $y^{2}=6x$,nearest to the point $\left(3, \frac{3}{2}\right)$ is $(\alpha, \beta)$,then $2(\alpha+\beta)$ is equal to $.....$

Find the vertex,focus,axis,length of latus rectum,and the equation of the directrix for the parabola $y^{2} = 4x + 4y$.

The ordinates of the vertices of a triangle inscribed in the parabola $y^2 = 4ax$ are $y_1, y_2, y_3$. Find the area of the triangle.

If the normal at the point $t_1$ (i.e.,at $(at_1^2, 2at_1)$) on the parabola $y^2 = 4ax$ meets the parabola again at the point $t_2$,then $t_1t_2$ is equal to:

Vedclass Test Series

Exam Paper Generator

Online Exam Module