JEE Mathematics 2023 Question with Solution

Given: The curves are $x = 2y^2$ and $x = 1 + y^2$. We need the distance of the point $\left(6, -2\sqrt{2}\right)$ from their common tangent.

Find: The perpendicular distance from the point to the common tangent line.

From the solution, the common tangent is obtained as

$$x - 2\sqrt{2}y + 1 = 0$$

This is equivalent to

$$y = \frac{1}{2\sqrt{2}}x + \frac{1}{2\sqrt{2}}$$

so the required tangent is identified.

Now use the distance of a point $\left(x_1, y_1\right)$ from the line $ax + by + c = 0$:

$$d = \frac{|ax_1 + by_1 + c|}{\sqrt{a^2 + b^2}}$$

Here,

$$a = 1, \quad b = -2\sqrt{2}, \quad c = 1$$

and the point is

$$\left(x_1, y_1\right) = \left(6, -2\sqrt{2}\right)$$

Substituting,

$$d = \frac{|6 - 2\sqrt{2}(-2\sqrt{2}) + 1|}{\sqrt{1 + 8}}$$

Since

$$-2\sqrt{2}(-2\sqrt{2}) = 8$$

we get

$$d = \frac{|6 + 8 + 1|}{3} = \frac{15}{3} = 5$$

Therefore, the distance is $5$ and the correct option is B.

Given: The extracted solution identifies the common tangent as $x - 2\sqrt{2}y + 1 = 0$.

Find: Distance from $\left(6, -2\sqrt{2}\right)$ to this line.

Write the line in the standard form

$$ax + by + c = 0$$

with

$$a = 1, \quad b = -2\sqrt{2}, \quad c = 1$$

Apply the point-to-line distance formula:

$$d = \frac{|ax_1 + by_1 + c|}{\sqrt{a^2 + b^2}}$$

Substitute $\left(x_1, y_1\right) = \left(6, -2\sqrt{2}\right)$:

$$d = \frac{|1\cdot 6 + (-2\sqrt{2})(-2\sqrt{2}) + 1|}{\sqrt{1^2 + (-2\sqrt{2})^2}}$$

Now simplify:

$$(-2\sqrt{2})(-2\sqrt{2}) = 8, \quad (-2\sqrt{2})^2 = 8$$

Hence,

$$d = \frac{|6 + 8 + 1|}{\sqrt{9}} = \frac{15}{3} = 5$$

Therefore, the required distance is $5$.