Find the vertex of the parabola whose focus is $(2,3)$. Also, $x-$axis is the tangent and $y-$axis is the normal to that parabola.

Question

Try no$(1):$ first wrote general 2nd degree curve and tried to subsitute the conditions such as $h^2 = ab$ where $ax^2 +by^2 + 2hxy +2gx+2fy+c=0$ is the general 2nd degree curve. Failed when tried to substitute tangent and normal conditons.

Try no$(2):$ wrote 2nd degree conic equation and then partially differentiated to get the centre of the conic but then failed to find the value of the vertex because of the denominator part missing. Kindly help me please!

Answer 1

One way to get such a parabola is to require the origin to be the tangent point to the $x$-axis, so that the $y$-axis automatically is normal there: Let $(x-2)^2+(y-3)^2=(ax+by+c)^2$ with $a^2+b^2=1$ be our equation. Then expanding we get:

$$(1-b^2)y^2-2abxy+(1-a^2)x^2+(-2ac-4)x+(-2bc-6)y+(13-c^2)=0$$

Now for the parabola to be tangent to the $x$-axis at the origin the constant term must vanish (since the origin is on the parabola) and the coefficient of $x$ must vanish since the tangent cone should be a multiple of $y$ alone (since $y=0$ defines the $x$-axis):

$$c^2=13, ac=-2\text{ remembering } a^2+b^2=1.$$

Analysing this and skipping the solution that leads to a degenerate conic we get

$$(x-2)^2+(y-3)^2=\frac{(3y-2x+13)^2}{13}$$ i.e. $(3x+2y)^2=156y$ and the vertex of this is the intersection with the axis $3x+2y=12$: $$(\frac{44}{13},\frac{12}{13}).$$

Edit: By @amd's observation we're done.

Answer 2

Instead of trying to beat the general conic equation into submission, take advantage of geometrical properties of parabolas.

If the $y$-axis is normal to the parabola, then its tangent at the point at which it intersects the $y$-axis must be parallel to the $x$-axis. However, you also know that the $x$-axis itself is a tangent, so the parabola must pass through the origin. You can now use the reflective property of parabolas to determine its axis direction: the reflection in the $y$-axis of the ray from the focus through the origin parallels the parabola’s axis. This gives $(-2,3)$ as the direction vector of the axis.

To find the directrix, you can use the fact that the foot of a perpendicular dropped from the focus to any tangent lies on the tangent to the parabola’s vertex. Since you know that the $x$-axis is tangent to the parabola, this means that $(2,0)$ lies on the tangent to its vertex. This line is parallel to the directrix, but only half as far from the focus, which should let you construct an equation for the directrix, after which you can use the definition of a parabola as the locus of points equidistant from a fixed point and line to construct its equation.