The equation of the lines joining the vertex | vedclass

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

Question

(D) The equation of the parabola is $y^2 = 6x$.Given the abscissa (x-coordinate) is $24$,we substitute $x = 24$ into the equation:$y^2 = 6(24) = 144$$

Vedclass · Accepted Answer

$2y \pm x = 0$ and $x \pm 2y = 0$

Vedclass · Answer

(D) The equation of the parabola is $y^2 = 6x$.Given the abscissa (x-coordinate) is $24$,we substitute $x = 24$ into the equation:$y^2 = 6(24) = 144$$y = \pm 12$.Thus,the points on the parabola are $(24, 12)$ and $(24, -12)$.The vertex of the parabola $y^2 = 6x$ is $(0, 0)$.The equation of the line joining $(0, 0)$ and $(24, 12)$ is $y - 0 = \frac{12 - 0}{24 - 0}(x - 0)$ $\Rightarrow y = \frac{1}{2}x$ $\Rightarrow 2y = x$.The equation of the line joining $(0, 0)$ and $(24, -12)$ is $y - 0 = \frac{-12 - 0}{24 - 0}(x - 0)$ $\Rightarrow y = -\frac{1}{2}x$ $\Rightarrow 2y = -x$.Combining these,we get $2y = \pm x$,which can also be written as $x \pm 2y = 0$.

The equation of the lines joining the vertex of the parabola $y^2 = 6x$ to the points on it whose abscissa is $24$ is:

Similar Questions

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.

What is the distance between the focus and the directrix of the parabola $x^2 = -8y$?

Find the area of the triangle formed by the points $(at_1^2, 2at_1)$,$(at_2^2, 2at_2)$,and $(at_3^2, 2at_3)$.

Find the minimum distance from the point $(0, c)$ to the parabola $y = x^2$,where $0 \leq c \leq 5$.

The curve described parametrically by $x = t^2 + t + 1$ and $y = t^2 - t + 1$ represents:

Vedclass Test Series

Exam Paper Generator

Online Exam Module