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Question

Mathematics Question on Hyperbola

Let P(x0,y0)P \left(x_0, y_0\right) be the point on the hyperbola 3x24y2=363 x^2-4 y^2=36, which is nearest to the line 3x+2y=13 x+2 y=1 Then 2(y0x0)\sqrt{2}\left(y_0-x_0\right) is equal to :

A

-9

B

3

C

9

D

-3

Answer

-9

Explanation

Solution

3x2−4y2=36
3x+2y=1
m=−32\frac 32
m=+3 sec θ12 tan θ\frac {3 \ sec \ θ}{\sqrt {12}\ tan\ θ}
⇒m = 312\frac {3}{12} × 1sin θ\frac{1}{sin\ θ​} = −32\frac 32
sinθ =−13\frac {1}{\sqrt 3}
(12\sqrt {12}​⋅sec θ, 3tan θ)
(12\sqrt {12}​⋅32\frac {\sqrt 3}{\sqrt 2}​​, −3×12\frac {1}{\sqrt 2}​) ⇒ (62\frac {6}{\sqrt 2}​, ​−32\frac {3}{\sqrt 2}​)
Now,
2(y0x0)\sqrt 2 (y_0 - x_0) = 2\sqrt 2 (- 32\frac {3}{\sqrt 2} - 62\frac {6}{\sqrt 2} ) = - 3 - 6 = - 9

So, the correct option is (A): - 9