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Q.
The coordinates of a point on the hyperbola $\frac{x^{2}}{24}-\frac{y^{2}}{18}=1$ which is nearest to the line $3 x+2 y+1=0$ are
Conic Sections
Solution:
Point $P$ is nearest to the given line if the tangent at $P$ is parallel to the given line.
Now, the slope of tangent at $P\left(x_{1}, y_{1}\right)$ is
$\left(\frac{d y}{d x}\right)_{\left(x_{1}, y_{1}\right)}=\frac{18 y_{1}}{24 x_{1}}=\frac{3}{4} \frac{y_{1}}{x_{1}} $
which must be equal to $-3 / 2 $.
Therefore,
$ \frac{3}{4} \frac{y_{1}}{x_{1}}=-\frac{3}{2}$
or $ y_{1}=-2 x_{1} $.... (i)
Also, $\left(x_{1}, y_{1}\right)$ lies on the curve.
Hence, $ \frac{x_{1}^{2}}{24}-\frac{y_{1}^{2}}{18}=1$ (ii)
Solving (i) and (ii), we get two points $(6,-3)$ and $(-6,3)$ of which $(6,-3)$ is the nearest.
