JEE Mathematics 2023 Question with Solution

Given: The point $(3, \alpha)$ lies on the parabola $y^2 = 12x$, and the tangent there is perpendicular to the line $2x + 2y = 3$.

Find: The square of the distance of the point $(6, -4)$ from the normal to the hyperbola $\alpha^2 x^2 - 9y^2 = 9\alpha^2$ at the point $(\alpha - 1, \alpha + 2)$.

Since $(3, \alpha)$ lies on $y^2 = 12x$,

$$\alpha^2 = 12 \cdot 3 = 36$$

So,

$$\alpha = \pm 6$$

For the parabola $y^2 = 12x$,

$$2y \frac{dy}{dx} = 12$$

Hence the slope of the tangent is

$$\frac{dy}{dx} = \frac{6}{\alpha}$$

The line $2x + 2y = 3$ has slope $-1$, so a perpendicular tangent must have slope $1$. Therefore,

$$\frac{6}{\alpha} = 1$$

which gives

$$\alpha = 6$$

Thus the hyperbola becomes

$$\frac{x^2}{9} - \frac{y^2}{36} = 1$$

Also, the given point on it is

$$(\alpha - 1, \alpha + 2) = (2, 8)$$

Using the extracted working, the normal is taken as

$$2x + 5y - 50 = 0$$

Now the distance of $(6, -4)$ from this line is

$$\frac{|2(6) + 5(-4) - 50|}{\sqrt{2^2 + 5^2}}$$ $$= \frac{|12 - 20 - 50|}{\sqrt{29}} = \frac{58}{\sqrt{29}}$$

Therefore, the square of the distance is

$$\left(\frac{58}{\sqrt{29}}\right)^2 = \frac{3364}{29} = 116$$

So the required answer is $116$.

Given: the solution concludes with Correct Answer: $116$ and shows the computation of the squared distance.

Find: The numerical value to be filled in the blank.

From the provided working:

1. $\alpha^2 = 36$, so $\alpha = \pm 6$.
2. Perpendicularity with the line $2x + 2y = 3$ gives tangent slope $1$, hence $\alpha = 6$.
3. The normal used in the working is

$$2x + 5y - 50 = 0$$

4. Distance of $(6, -4)$ from this line is

$$\frac{|2(6) + 5(-4) - 50|}{\sqrt{2^2 + 5^2}} = \frac{58}{\sqrt{29}}$$

5. Squaring,

$$\frac{3364}{29} = 116$$

Therefore, the blank should be filled with $116$.