Q.Find the equation of the parabola which is symmetric about the $y$-axis, and passes through the point $(2, -3)$.

Concept & Intuition

We need the locus of $P(x,y,z)$ for which

• The X-axis is the set of points $(t,0,0)$, so the perpendicular distance from $P$ to it is the distance measured in the $y$-$z$ directions: $d_X=\sqrt{y^2+z^2}$.
• The YZ-plane is $x=0$, so the perpendicular distance from $P$ to it is $d_{YZ}=|x|$.

Step-by-Step Derivation

1. Write the two distances.

   $d_X=\sqrt{y^2+z^2}$ and $d_{YZ}=|x|$.

2. Impose the ratio.

3. Square both sides (both are non-negative, so no extraneous roots):

4. Rearrange to standard form. $4x^2-9y^2-9z^2=0.$ …

Shift the origin. Put $x=X+h,\ y=Y+5$ into $y=x^3-9x^2+cx-d$ and require the result to be $Y=X^3$.

Vanishing $X^2$ term. The coefficient of $X^2$ is $3h-9=0$, so

Vanishing $X$ term. The coefficient of $X$ is $3h^2-18h+c=27-54+c=0$, so

Equivalently $x^3-9x^2+27x-d=(x-3)^3+(27-d)$, confirming the depressed-cubic shift is exact. …

Concept & Intuition

We have a hyperbola $\frac{x^2}{a^2} - \frac{y^2}{9} = 1$. Its foci are at $(\pm ae, 0)$ where $e = \sqrt{1 + \frac{b^2}{a^2}}$ and here $b^2 = 9$. The focus on the positive X‑axis is $S = (ae, 0)$.

We are given a point $P(\theta)$ on the hyperbola — this notation usually means the parametric form: $P = (a\sec\theta, 3\tan\theta)$.

We know $SP = 6$ and $SQ = \sqrt{26}$ with $Q = (0,1)$. The distance $SQ$ will let us find $ae$ (the x‑coordinate of S). Then using $SP = 6$ and the parametric coordinates, we can solve for $\theta$.

Step‑by‑step solution

1. Find the focus coordinate $S = (ae, 0)$ using $SQ = \sqrt{26}$. $Q = (0,1)$, so

Hence

So $S = (5, 0)$.

2. Relate $a$ and $e$. For a hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$, we have $e = \sqrt{1 + \frac{b^2}{a^2}}$. Here $b^2 = 9$, so

But we also have $ae = 5$. Substituting:

Squaring: $a^2 + 9 = 25 \Rightarrow a^2 = 16 \Rightarrow a = 4$ (positive).

Then $e = \frac{5}{a} = \frac{5}{4}$.

3. Parametric form of P. The hyperbola $\frac{x^2}{16} - \frac{y^2}{9} = 1$ has parametric coordinates

4. Use $SP = 6$.

Expand:

Recall $\tan^2\theta = \sec^2\theta - 1$. Substitute:

Simplify:

So

Divide by 5:

5. Solve the quadratic in $\sec\theta$. $\sec\theta = \frac{8 \pm \sqrt{64 + 80}}{10} = \frac{8 \pm \sqrt{144}}{10} = \frac{8 \pm 12}{10}.$ …

Step 1 — line in slope form. $x\cos\alpha+y\sin\alpha=2\sqrt{3}$ gives

Step 2 — tangency to $\dfrac{x^2}{16}+\dfrac{y^2}{8}=1$. Here $a^2=16,\,b^2=8$, and tangency requires $c^2=a^2m^2+b^2$:

Step 3 — solve. Using $\cot^2\alpha=\dfrac{1-\sin^2\alpha}{\sin^2\alpha}$ and multiplying by $\sin^2\alpha$: …