JEE Mathematics 2023 Question with Solution

Given: $P(h,k)$ lies on the parabola $x = 4y^2$ and is nearest to $Q(0,33)$.

Find: The distance of $P$ from the directrix of the parabola $y^2 = 4(x+y)$.

For the parabola $x = 4y^2$, write it as $y^2 = \frac{x}{4}$, so in the standard form $y^2 = 4ax$ we get $a = \frac{1}{16}$.

The normal to $y^2 = 4ax$ at parameter $t$ is

$$y = -tx + 2at + at^3$$

Substituting $a = \frac{1}{16}$,

$$y = -tx + \frac{t}{8} + \frac{t^3}{16}$$

Since this normal passes through $Q(0,33)$,

$$33 = \frac{t}{8} + \frac{t^3}{16}$$

Multiplying by $16$,

$$528 = 2t + t^3$$

So,

$$t^3 + 2t - 528 = 0$$

From the factorization shown in the solution,

$$(t-8)(t^2+8t+66)=0$$

Hence $t = 8$.

The point on the parabola is

$$P(at^2, 2at)$$

Therefore,

$$P = \left(\frac{1}{16}\times 64, 2\times \frac{1}{16}\times 8\right) = (4,1)$$

Now transform the second parabola:

$$y^2 = 4(x+y)$$ $$y^2 - 4y = 4x$$ $$(y-2)^2 = 4(x+1)$$

This is of the form $(y-k)^2 = 4a(x-h)$ with $a=1$ and vertex $(-1,2)$. Hence its directrix is

$$x = -2$$

So the distance of $P(4,1)$ from the directrix $x=-2$ is

$$|4-(-2)| = 6$$

Therefore, the distance is $6$, so the correct option is D.

Given: $x = 4y^2$ and nearest point from $Q(0,33)$.

Find: Distance from that point to the directrix of $y^2 = 4(x+y)$.

The solution states the normal equation and then uses the condition that it passes through $Q(0,33)$. While one displayed line in the first approach writes $33 = \frac{t^2}{16} + \frac{t^3}{16}$, the second approach and the subsequent factorization clearly support the corrected working

$$33 = \frac{t}{8} + \frac{t^3}{16}$$

which gives

$$t^3 + 2t - 528 = 0$$

and hence $t=8$.

Then

$$P(at^2,2at) = (4,1)$$

Also,

$$y^2 = 4(x+y) \Rightarrow (y-2)^2 = 4(x+1)$$

So the directrix is

$$x=-2$$

Finally,

$$\text{Distance} = |4-(-2)| = 6$$

Thus, despite a minor inconsistency in one intermediate printed expression, the extracted working leads to the final answer $6$ and hence option D.