JEE Mathematics 2023 Question with Solution

Given: The hyperbola is

$$3x^2 - 4y^2 = 36$$

and the line is

$$3x + 2y = 1.$$

Find: The value of $\sqrt{2}(y_0 - x_0)$ where $P(x_0,y_0)$ is the nearest point on the hyperbola to the given line.

From the solution working, the slope of the line is

$$m = -\frac{3}{2}.$$

For a point on the hyperbola, the solution uses the parametric form

$$\left(\sqrt{12}\sec\theta,\,3\tan\theta\right).$$

The slope condition for the nearest point is written as

$$\frac{3}{\sqrt{12}}\times\frac{1}{\sin\theta} = -\frac{3}{2}.$$

So,

$$\sin\theta = -\frac{1}{\sqrt{3}}.$$

Hence the corresponding point is

$$\left(\sqrt{12}\cdot \frac{\sqrt{3}}{2},\,-3\cdot \frac{1}{\sqrt{2}}\right)$$

which simplifies to

$$\left(\frac{6}{\sqrt{2}},\,-\frac{3}{\sqrt{2}}\right).$$

Therefore,

$$\sqrt{2}\left(y_0-x_0\right)=\sqrt{2}\left(-\frac{3}{\sqrt{2}}-\frac{6}{\sqrt{2}}\right)=-9.$$

So the computed value is $-9$.

However, the solution explicitly states "The Correct Option is A", while the computed value matches option C. for the answer field, the marked answer is A even though the working gives $-9$.

Given:

$$3x^2 - 4y^2 = 36, \qquad 3x + 2y = 1$$

Find: $\sqrt{2}(y_0-x_0)$.

The solution states:

1. Slope of the line is

$$m=-\frac{3}{2}.$$

2. The nearest point condition is taken as

$$m=\pm \frac{\sec\theta \cdot 3}{\sqrt{12}\cdot \tan\theta}.$$

3. Equating slopes,

$$\frac{3}{\sqrt{12}}\times \frac{1}{\sin\theta}=-\frac{3}{2}.$$

4. Therefore,

$$\sin\theta=-\frac{1}{\sqrt{3}}.$$

5. The corresponding point is

$$\left(\sqrt{12}\sec\theta,\,3\tan\theta\right).$$

6. Substituting the trigonometric values,

$$\left(\sqrt{12}\cdot \frac{\sqrt{3}}{2},\,-3\cdot \frac{1}{\sqrt{2}}\right) =\left(\frac{6}{\sqrt{2}},\,-\frac{3}{\sqrt{2}}\right).$$

7. Now evaluate the asked expression:

$$\sqrt{2}(y_0-x_0)=\sqrt{2}\left(-\frac{3}{\sqrt{2}}-\frac{6}{\sqrt{2}}\right)=-9.$$

Thus the final value obtained from the working is $-9$.