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  • The coordinates of a point on the hyperbola <img alt="\mathrm{\frac{x^2}{24}-\frac{y^2}{18}=1}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmathrm%7B%5Cfrac%7Bx%5E2%7D%7B24%7D-%5Cfrac%7

The coordinates of a point on the hyperbola \mathrm{\frac{x^2}{24}-\frac{y^2}{18}=1} which is nearest to the line \mathrm{3 x+2 y+1=0 }are
 

Option: 1

 (6,3)
 


Option: 2

 (-6,-3)
 


Option: 3

 (-6,3)
 


Option: 4

 (6,-3)


Answers (1)

best_answer

Point P is is nearest to the given line if the tangent at P is parallel to the given line.
Now, the slope of tangent at \mathrm{P\left(x_1, y_1\right)} is \mathrm{\left(\frac{d y}{d x}\right)_{\left(x_1, y_1\right)}=\frac{18 x_1}{24 y_1}=\frac{3}{4} \frac{x_1}{y_1}} which must be equal to \mathrm{-\frac{3}{2}}

Therefore,

\mathrm{\frac{3}{4} \frac{x_1}{y_1}=-\frac{3}{2} }

or \mathrm{x_1=-2 y_1 \quad \quad \dots(1)}
Also, \mathrm{\left(x_1, y_1\right)} lies on the curve. Hence,

\mathrm{\frac{x_1^2}{24}-\frac{y_1^2}{18}=1 \quad \quad \dots(2)}
Solving (1) and (2), we get two points (6,-3) and (-6,3) of which (-6,3) is the nearest.

Posted by

HARSH KANKARIA

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