If the axis of a parabola is horizontal and i | vedclass

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

Question

(D) The general equation of a parabola with a horizontal axis is $x = ay^2 + by + c$.Since the parabola passes through $(0, 0)$,we have $0 = a(0)^2 +

Vedclass · Accepted Answer

None of these

Vedclass · Answer

(D) The general equation of a parabola with a horizontal axis is $x = ay^2 + by + c$.Since the parabola passes through $(0, 0)$,we have $0 = a(0)^2 + b(0) + c$,which implies $c = 0$.So,the equation is $x = ay^2 + by$.Using the point $(0, -1)$,we get $0 = a(-1)^2 + b(-1)$,which means $a - b = 0$,so $a = b$.Using the point $(6, 1)$,we get $6 = a(1)^2 + b(1)$,which means $a + b = 6$.Substituting $a = b$ into $a + b = 6$,we get $2a = 6$,so $a = 3$ and $b = 3$.The equation is $x = 3y^2 + 3y$,or $3y^2 + 3y - x = 0$.Comparing this with the given options,none of the provided equations match the result.

If the axis of a parabola is horizontal and it passes through the points $(0, 0), (0, -1)$ and $(6, 1)$,then its equation is

Similar Questions

The parabolas $ax^2 + 2bx + cy = 0$ and $dx^2 + 2ex + fy = 0$ intersect on the line $y = 1$. If $a, b, c, d, e, f$ are positive real numbers and $a, b, c$ are in $G.P.$,then

The area of the triangle formed by the lines joining the vertex of the parabola $x^2 = 20y$ to the ends of its latus rectum is

If the chord joining the points $(at_1^2, 2at_1)$ and $(at_2^2, 2at_2)$ of the parabola $y^2 = 4ax$ passes through the focus of the parabola,then

The equation of the parabola with $(-3, 0)$ as focus and $x + 5 = 0$ as directrix is

Find the equation of the parabola that satisfies the following conditions: Vertex $(0, 0)$,focus $(-2, 0)$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module