If {x^2} + 6x + 20y - 51 = 0,then the axis of | vedclass

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

Question

(A) Given equation of the parabola is ${x^2} + 6x + 20y - 51 = 0$. Rearranging the terms,we get: ${x^2} + 6x = -20y + 51$. Completing the square on th

Vedclass · Accepted Answer

$x + 3 = 0$

Vedclass · Answer

(A) Given equation of the parabola is ${x^2} + 6x + 20y - 51 = 0$. Rearranging the terms,we get: ${x^2} + 6x = -20y + 51$. Completing the square on the left side: ${x^2} + 6x + 9 = -20y + 51 + 9$. This simplifies to: ${(x + 3)^2} = -20y + 60$. Factoring out $-20$ on the right side: ${(x + 3)^2} = -20(y - 3)$. Comparing this with the standard form ${(x - h)^2} = 4a(y - k)$,the axis of the parabola is given by $x - h = 0$. Here,$h = -3$,so the axis is $x + 3 = 0$.

If ${x^2} + 6x + 20y - 51 = 0$,then the axis of the parabola is

Similar Questions

The ordinates of the points $P$ and $Q$ on the parabola with focus $(3,0)$ and directrix $x = -3$ are in the ratio $3:1$. If $R(\alpha, \beta)$ is the point of intersection of the tangents to the parabola at $P$ and $Q$,then $\frac{\beta^2}{\alpha}$ is equal to $.............$.

If the vertices of a triangle are $(am_1^2, 2am_1), (am_2^2, 2am_2),$ and $(am_3^2, 2am_3),$ then the area of the triangle is

The length of the focal chord of the parabola $y^2 = 4ax$ at a distance $p$ from the vertex is:

The parabola $y^2 = x$ is symmetric about

The equation of the axis of the parabola $2x^2 + 5y - 3x + 4 = 0$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module