What is the length of the latus rectum of the | vedclass

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

Question

(B) The given equation is $x = ay^2 + by + c$.We can rewrite this by completing the square for $y$:$x = a(y^2 + \frac{b}{a}y) + c$$x = a(y^2 + \frac{b

Vedclass · Accepted Answer

$\frac{1}{|a|}$

Vedclass · Answer

(B) The given equation is $x = ay^2 + by + c$.We can rewrite this by completing the square for $y$:$x = a(y^2 + \frac{b}{a}y) + c$$x = a(y^2 + \frac{b}{a}y + \frac{b^2}{4a^2}) + c - \frac{b^2}{4a}$$x - (c - \frac{b^2}{4a}) = a(y + \frac{b}{2a})^2$$(y + \frac{b}{2a})^2 = \frac{1}{a}(x - (c - \frac{b^2}{4a}))$This is in the standard form $(y - k)^2 = 4p(x - h)$,where $4p = \frac{1}{a}$.The length of the latus rectum is given by $|4p|$.Therefore,the length is $|\frac{1}{a}| = \frac{1}{|a|}$.

What is the length of the latus rectum of the parabola $x = ay^2 + by + c$?

Similar Questions

Consider the conic $C: 25(x - 1)^2 + 25(y + 1)^2 = (3x - 4y)^2$. If the curve $E$ is the locus of the point of intersection of perpendicular tangents to the conic $C$,then the minimum distance between the curve $E$ and the point $(2, -1)$ is:

Let $P(\alpha, \beta)$ be a point on the parabola $y^2 = 4x$ which is at minimum distance from the circle $x^2 + y^2 - 4x - 20y + 103 = 0$. Then $\alpha \beta$ is

The locus of the midpoints of the chords of the parabola $y^2 = 4ax$ which pass through the origin is:

If the parabolas $y^2 = 4b(x - c)$ and $y^2 = 8ax$ have a common normal,then which one of the following is a valid choice for the ordered triad $(a, b, c)$?

If the tangent drawn at the point $P(4,8)$ to the parabola $y^2=16x$ meets the parabola $y^2=16x+80$ at $A$ and $B$,then the mid-point of $AB$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module