The coordinates of the focus of the parabola | vedclass

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

Question

(A) Given parametric equations are $x=5t^2+2$ and $y=10t+4$.From the second equation,$t = \frac{y-4}{10}$.Substituting $t$ into the first equation:$x-

Vedclass · Accepted Answer

$(7,4)$

Vedclass · Answer

(A) Given parametric equations are $x=5t^2+2$ and $y=10t+4$.From the second equation,$t = \frac{y-4}{10}$.Substituting $t$ into the first equation:$x-2 = 5\left(\frac{y-4}{10}\right)^2 = 5\left(\frac{(y-4)^2}{100}\right) = \frac{(y-4)^2}{20}$.Thus,$(y-4)^2 = 20(x-2)$.Comparing this with the standard form $(y-k)^2 = 4a(x-h)$,we have $h=2, k=4$,and $4a=20$,which gives $a=5$.The focus of the parabola $(y-k)^2 = 4a(x-h)$ is given by $(h+a, k)$.Substituting the values,the focus is $(2+5, 4) = (7,4)$.

The coordinates of the focus of the parabola described parametrically by $x=5t^2+2, y=10t+4$ (where $t$ is a parameter) are

Similar Questions

Let $M$ be the foot of the perpendicular from a point $P$ on the parabola $y^2=8(x-3)$ onto its directrix and let $S$ be the focus of the parabola. If $\triangle SPM$ is an equilateral triangle,then $P$ is equal to

The point on the parabola $y^2 = 18x$,for which the ordinate is three times the abscissa,is

The radius of the smallest circle which touches the parabolas $y = x^2 + 2$ and $x = y^2 + 2$ is

Find the equation of the parabola whose focus is $(-1, -2)$ and whose directrix is the line $x - 2y + 3 = 0$.

The maximum number of common normals of $y^2=4ax$ and $x^2=4by$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module