Question

a parabola is drawn with focus at 4,6 and vertex at focus of parabola y²-16x-6y+9=0 the equation of the parabola is​

Answer 1

Given:

Equation of a Parabola:

$y^2-16x-6y+9=0$

Another parabola with Focus at $F_2(4, 6)$

Vertex of other parabola is at focus of given parabola.

To find:

Equation of other parabola.

Solution:

We can see that the given equation is a quadratic in $y$.

Therefore, it is a horizontal parabola.

The given equation can be re-written as:

$y^2-6y+9=16x\\\Rightarrow (y-3)^2=16(x-0)$

Comparing with standard equation of a horizontal parabola:

$(y-k)^2=4p(x-h)$

We get,

$h=0, k=3, p = 4$

Formula for focus of a horizontal parabola is $(h+p, k)$

Therefore $F_1(0+4, 3)\Rightarrow F_1(4, 3)$

As per given statement:

$F_1$ is vertex of other parabola.

We can say that,

Vertex of parabola is (4, 3) and focus of parabola is (4, 6)

There is change in y coordinate in vertex and focus.

Therefore, it is a vertical parabola.

If vertex is $(h, k)$ , then focus is at $(h, k+p)$.

$6=3+p\\\Rightarrow p = 3$

Equation for a vertical parabola is given as:

$(x-h)^2=4p(y-k)$

Putting the values, we get:

$(x-4)^2=4\times 3(y-3)\\\Rightarrow x^{2} -8x-12y+52=0$

Hence, the equation of parabola is:

$x^{2} -8x-12y+52=0$