JEE 2023 — Mathematics PYQ

The point on a curve nearest to a line is the point where the tangent is parallel to that line.

1. Find the slope of the given line:

The line is $3x+2y=1$, so its slope $m=-\frac{3}{2}$.

2. Differentiate the hyperbola equation to find the slope of the tangent:

Differentiating with respect to $x$:

Set this equal to the slope of the line:

3. Substitute $x_0=-2y_0$ into the hyperbola equation:

4. Find the corresponding $x_0$:

• If $y_0=\frac{3}{\sqrt{2}}$, then $x_0=-2(\frac{3}{\sqrt{2}})=-\frac{6}{\sqrt{2}}$.

• If $y_0=-\frac{3}{\sqrt{2}}$, then $x_0=\frac{6}{\sqrt{2}}$.

5. Determine the nearest point:

Check the distance of $P(x_0,y_0)$ from $3x+2y-1=0$.

For $P(-\frac{6}{\sqrt{2}},\frac{3}{\sqrt{2}})$: Distance $\approx |3(-\frac{6}{\sqrt{2}})+2(\frac{3}{\sqrt{2}})-1|=|-\frac{12}{\sqrt{2}}-1|=6\sqrt{2}+1$.

For $P(\frac{6}{\sqrt{2}},-\frac{3}{\sqrt{2}})$: Distance $\approx |3(\frac{6}{\sqrt{2}})+2(-\frac{3}{\sqrt{2}})-1|=|\frac{12}{\sqrt{2}}-1|=6\sqrt{2}-1$.

The nearest point is $x_0=\frac{6}{\sqrt{2}}$ and $y_0=-\frac{3}{\sqrt{2}}$.

6. Calculate the final value:

Correct Option: (1)