Q.If $x = t^2$, $y = t^3$, then $\dfrac{d^2 y}{dx^2}$ is
(A) $\dfrac{3}{2}$
(B) $\dfrac{3}{4t}$
(C) $\dfrac{3}{2t}$
(D) $\dfrac{3}{4}$

The problem gives a parabola in parametric equations:

$x = -2 + 2t^2$,

$y = 2 + 4t$.

We need the product $\alpha\beta$, where $t = \alpha$ and $t = \beta$ are the parameters at the ends of the latus rectum.

1. Recognise the standard form.

   The parametric form $x = a t^2$, $y = 2 a t$ describes the standard parabola $y^2 = 4 a x$, opening to the right. Here we have shifts: $x = -2 + 2t^2$ and $y = 2 + 4t$. Compare:

   $2t^2$ suggests $a = 2$ (since $x = a t^2$ gives coefficient $a$), and $4t$ matches $2 a t$ because $2 \cdot 2 \cdot t = 4t$. So the underlying parabola is $y^2 = 8x$ (since $4a = 8$), shifted so that its vertex is at $(-2, 2)$.

2. What is the latus rectum?

   For any parabola, the latus rectum is the chord through the focus, perpendicular to the axis. For $y^2 = 4ax$, the focus is at $(a, 0)$, and the ends of the latus rectum are at $(a, \pm 2a)$. In parametric form $x = a t^2$, $y = 2 a t$, these ends correspond to $t = \pm 1$.

   Why? Because at $t = 1$, $x = a$, $y = 2a$; at $t = -1$, $x = a$, $y = -2a$. So the two parameter values are $1$ and $-1$, and their product is $-1$.

3. Apply to the shifted parabola. …

Concept and Intuition

A normal to a parabola is a line perpendicular to the tangent at the point of contact. For the parabola $y^2 = 4ax$, the equation of the normal at a point with parameter $t$ (or slope $m$) is well-known. When we ask "how many normals pass through a given external point?", we are essentially solving for the slope(s) $m$ that satisfy the normal equation when the point coordinates are plugged in. This leads to a cubic equation in $m$. The number of real normals is the number of real roots of that cubic. A cubic always has at least one real root, so the answer is never zero — but it can be one, two, or three depending on the point's location relative to the parabola.

Here, the point is $(2,0)$ and the parabola is $y^2 = 7x$. We'll derive the cubic and count its real roots.

Step-by-step solution

1. Identify the parabola's standard form The given parabola is $y^2 = 7x$. Compare with $y^2 = 4ax$:

2. Equation of the normal in slope form For $y^2 = 4ax$, the normal at a point with slope $m$ (where $m \neq 0$) is:

This is a standard result derived from the parametric form $(at^2, 2at)$ with $m = -1/t$.

Substituting $a = \frac{7}{4}$:

3. Impose the condition that the normal passes through $(2,0)$ Plug $x = 2$, $y = 0$:

Simplify:

4. Factor the cubic equation Multiply through by 4 to avoid fractions:

Factor out $m$:

5. Find the real roots

   • $m = 0$ is one real root.
   • $7m^2 + 6 = 0$ gives $m^2 = -\frac{6}{7}$, which has no real solutions.

   So there is exactly one real value of $m$.

6. Interpretation …