Parabola Property Problem related to Focus and its images through tangents

Question

A parabola $S = 0$ has its vertex at $\left( { - 9,3} \right)$ and it touches the x-axis at the origin then equation of axis of symmetry of the aforesaid parabola can be

(A) $x-y+12=0$

(B) $x-2y+15=0$

(C) $2x-y+21=0$

(D) $x+y+6=0$

My approach is as follow

Note: The image of Focus to any Tangent lines lies on Directrix.

The parabola touching the x-axis means one of the tangent ${T_1}:y = 0$.

Note: The tangents at the extremities of a focal chord of a parabola are perpendicular to each other and intersect at the directrix of the parabola.

Hence as per the figure the line BS is ${T_1}:y = 0$ and AS is ${T_2}:x = k$ are perpendicular tangents.

Let the focus be $F\left( {\alpha ,\beta } \right)$ , therefore $B'\left( {\alpha , - \beta } \right)$ is image of ${T_1}:y = 0$

Let the focus be $F\left( {\alpha ,\beta } \right)$ , therefore $A'\left( {2k - \alpha ,\beta } \right)$ is image of ${T_2}:x = k$

Therefore the point $S\left( { - 18 - \alpha ,6 - \beta } \right)$ where the axis of symmetry intersect the directrix as $V(9,3)$ is the vertex of the parabola.

Therefore $A'\left( {2k - \alpha ,\beta } \right)$, $S\left( { - 18 - \alpha ,6 - \beta } \right)$ and $B'\left( {\alpha , - \beta } \right)$ are collinear.

Using Area Property for collinear lines

$ \Rightarrow 144 + 12\alpha - 36\beta - 4\alpha \beta - \left( {144 + 12k} \right) = 0 \Rightarrow 12\alpha - 36\beta - 4\alpha \beta - 12k = 0 \Rightarrow 3\alpha - 9\beta - \alpha \beta - 3k = 0$

Not able to proceed

Answer 1

Let $T=(x_0,0)$ be the point where the tangent at vertex $V=(-9,3)$ intersects the x-axis and $P$ the point where the axis of the parabola intersects the x-axis (see figure below). Then we have $OT=PT$ ^(*) and therefore $P=(2x_0,0)$.

But $VT$ is perpendicular to $VP$: if $H=(-9,0)$ is the projection of $V$ on x-axis we have then $VH^2=PH\cdot TH$, that is: $$ (-9-2x_0)(x_0+9)=9 $$ which gives $x_0=-6$ or $x_0=-7.5$.

You can now easily find the two possible equations of axis $VP$.

^(*) Because, if $M$ is the midpoint of $OV$ we know that $MT\parallel PV$.

Answer 2

Secondary approach:

Let's assume the axis of symmetry as $\ x - ay + k_1 = 0$ and the tangent at the vertex as $ax + y + k_2 = 0$. Given that both lines are perpendicular and intersect at $(-9, 3)$, we can find $k_2$ in terms of $k_1$. Also, from the intersection point, we obtain the relationship $a = \frac{{k_1 - 9}}{3}$, which will be useful later. Additionally, we find $k_2 = 3k_1 - 30$.

Now, the general equation of a parabola (oblique or standard) is often written as: $$PM^2 = 4b PN$$ where $PM$ is the equation of the axis of symmetry, and $PN$ is the tangent at the vertex. This can be derived using $y^2 = 4ax$ expressed in terms of distances from the given lines.

Therefore, the equation of the parabola is:

$$ (x - ay + k_1)^2 = 4b(ax + y + 3k_1 - 30) $$

Since the point $(0,0)$ lies on the parabola, we have $$k_1^2 = 4b(3k_1 - 30).$$

Moreover, the line $y = 0$ is tangent to the parabola. Substituting this into the equation, we get:

$$ x^2 + x(2k_1 - 4ab) = 0 $$

For the equation to have a single root at $(0,0)$, the x coefficient must be $0$. Thus, we obtain $$2k_1 = 4ab.$$

Now, the first equation is: $$a = \frac{{k_1 - 9}}{3}.$$

The second equation is: $$k_1^2 = 4b(3k_1 - 30).$$

The third equation is: $$2k_1 = 4ab.$$

Now, we have three equations, and by eliminating variables, we can derive a single quadratic equation in terms of $k_1$. Solving this quadratic equation yields $k_1 = 15$ and $k_1 = 12$. Substituting these values back into the first equation gives $a = 2$ and $a = 1$, respectively. Therefore, the equations of the axis of symmetry are $$x - 2y + 15 = 0$$ or $$x - y + 12 = 0$$The entire equation of the parabola can be found by determining the value of $b$ and substituting it. Thanks!

Answer 3

Let the slope of tangent at vertex $V(-9,3)$ be $m$. Then the intercept $P$ of this tangent on the x-axis will be $\left(-\dfrac{9m+3}{m},0 \right)$

It is well known that the median of $\triangle VPO $ through $P$ will be parallel to the axis i.e. perpendicular to the tangent at the vertex.

So we get the condition $\dfrac{\dfrac{3}{2}}{\dfrac{9m+3}{m}-\dfrac{9}{2}}.m =-1$ which has solutions $m=-1,-2$ so that the axis can have slope $1$ or $\dfrac{1}{2}$

Hence the equation of the axis can be $x-y+12=0$ or $x-2y+15=0$