The parametric equations of the parabola x^2- | vedclass

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

Question

(A) Given equation of parabola is $x^2-8 x+12 y+15=0$. Completing the square for $x$: $x^2-8 x+16 = -12 y - 15 + 16$ $(x-4)^2 = -12 y + 1$ $(x-4)^2 =

Vedclass · Accepted Answer

$x=4+6 t, y=\frac{1}{12}-3 t^2$

Vedclass · Answer

(A) Given equation of parabola is $x^2-8 x+12 y+15=0$. Completing the square for $x$: $x^2-8 x+16 = -12 y - 15 + 16$ $(x-4)^2 = -12 y + 1$ $(x-4)^2 = -12(y - \frac{1}{12})$. Comparing this with the standard form $(x-h)^2 = -4a(y-k)$,we get $h=4$,$k=\frac{1}{12}$,and $4a=12$,so $a=3$. The parametric equations for $(x-h)^2 = -4a(y-k)$ are $x = h + 2at$ and $y = k - at^2$. Substituting the values,we get $x = 4 + 2(3)t = 4 + 6t$ and $y = \frac{1}{12} - 3t^2$. Thus,the correct option is $A$.

The parametric equations of the parabola $x^2-8 x+12 y+15=0$ are

Similar Questions

Find the locus of the midpoints of all focal chords of the parabola $y^2 = 4ax$.

What is the common tangent to the parabola $y^{2} = 8ax$ and the circle $x^{2} + y^{2} = 2a^{2}$?

If the line $y=2x+k$ is a normal to the parabola $y^2=4x$,then $k=$

Find the coordinates of the focus,axis,the equation of the directrix,and the length of the latus rectum of the parabola $y^{2}=8x$.

The equation of the latus rectum of a parabola is $x+y=8$ and the equation of the tangent at the vertex is $x+y=12$. Then the length of the latus rectum is

Vedclass Test Series

Exam Paper Generator

Online Exam Module