The point on the hyperbola 3x²-4y²=72 which is nearest to the line 3x+2y+1=0 is

Correct Answer is : (−6,3)(-6,3)(−6,3)

Correct Answer is : (−6,3)(-6,3)(−6,3)

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KEAM 2011 Math Paper

Question : 55 of 120

Marks: +1, -0

3x^2-4y^2=72

3x+2y+1=0

$(-6,3)$
$(6,3)$
$(-6,-3)$
$(6,-3)$
$(\sqrt{24},0)$

Solution:

First we check which point satisfy the equation of hyperbola. All points in options are satisfied the equation of hyperbola

3 x^{2}-4 y^{2}=72

Now, we find one-by-one the length of perpendicular from point on Ellipse to the line

3 x+2 y+1=0

P_{(-6,3)} =\; \frac{11}{\sqrt{13}}

P_{(6,3)} =\; \frac{25}{\sqrt{13}}

P_{(-6,-3)} =\; \frac{23}{\sqrt{13}}

P_{(6,-3)} =\; \frac{13}{\sqrt{13}}

P_{(\sqrt{24}, 0)} =\; \frac{3 \sqrt{24} + 1}{\sqrt{13}}

The minimum length is

P_{(-6,3)} .

So, the point

(-6,3)

is nearest to the given line.

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