Find the equation of the hyperbola whose, Focus is (2, 1) directrix is 2x+3y=1 and eccentricity = 2

Let (2, 1) be the focus and P (x, y) be a point on the hyperbola, Draw PM perpendicular from P on the directrix, Then, by definition sP=ePM ^ 2 =e^ 2 PM^ 2 (x-2)^ 2 +(y-1)^ 2 =2^ 2 [ 2x+3y-1 2^ 2 +3^ 2 ]^ 2 ^ 2 +4-4x+y^ 2 +1+2y= 4[2x+3y-1]^ 2 13 13[x^ 2 +y^ 2 -4x+2y+5]=4(2x+3y-1)^ 2 13x^ 2 +13y^ 2 -52x+26y+65=4[2x+3y-1]^ 2 13x^ 2 +13y^ 2 -52x+26y+65 =4 [(2x)^ 2 +(3y)^ 2 +(-1)^ 2 +2 2x 3y (-1)+2 (-1) 2x ] 13x^ 2 +13y^ 2 -52x+26y+65=4 [4x^ 2 +9y^ 2 +1+12xy-6y-4x ] 13x^ 2 +13y^ 2 -52x+26y+65=16x^ 2 +36y^ 2 +4+48xy-24y-16x 16x^ 2 -13 x^ 2 +36 y^ 2 -13y^ 2 +48xy-16x+52x-24y-26y+4-65=0 3x^ 2 +23y^ 2 +48+36x-50y-61=0 This is the required equation of the hyperbola.

Find the equation of the hyperbola whose, Focus is (2, 1) directr…

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Question

Find the equation of the hyperbola whose,
Focus is (2, 1) directrix is $2\text{x}+3\text{y}=1$ and eccentricity = 2

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Answer

Let (2, 1) be the focus and P (x, y) be a point on the hyperbola, Draw PM perpendicular from P on the directrix, Then, by definition
$\text{sP}=\text{ePM}$
$\Rightarrow\text{sP}^{2}=\text{e}^{2}\text{PM}^{2}$
$\Rightarrow(\text{x-2)}^{2}+(\text{y}-1)^{2}=2^{2}\Bigg[\frac{2\text{x}+3\text{y}-1}{\sqrt{2^{2}+3^{2}}}\Bigg]^{2}$
$\Rightarrow\text{x}^{2}+4-4\text{x}+\text{y}^{2}+1+2\text{y}=\frac{4[2\text{x}+3\text{y}-1]^{2}}{13}$
$\Rightarrow13[\text{x}^{2}+\text{y}^{2}-4\text{x}+2\text{y}+5]=4(2\text{x}+3\text{y}-1)^{2}$
$\Rightarrow13\text{x}^{2}+13\text{y}^{2}-52\text{x}+26\text{y}+65=4[2\text{x}+3\text{y}-1]^{2}$
$\Rightarrow13\text{x}^{2}+13\text{y}^{2}-52\text{x}+26\text{y}+65\\=4\Big[(2\text{x)}^{2}+(3\text{y})^{2}+(-1)^{2}+2\times2\text{x}\times3\text{y}\times(-1)+2\times(-1)\times2\text{x}\Big]$
$\Rightarrow13\text{x}^{2}+13\text{y}^{2}-52\text{x}+26\text{y}+65=4\Big[4\text{x}^{2}+9\text{y}^{2}+1+12\text{xy}-6\text{y}-4\text{x}\Big]$
$\Rightarrow13\text{x}^{2}+13\text{y}^{2}-52\text{x}+26\text{y+65}=16\text{x}^{2}+36\text{y}^{2}+4+48\text{xy}-24\text{y}-16\text{x}$
$\Rightarrow16\text{x}^{2}-13{\text{x}^{2}+36}\text{y}^{2}-13\text{y}^{2}+48\text{xy}-16\text{x}+52\text{x}-24\text{y}-26\text{y}+4-65=0$
$\Rightarrow3\text{x}^{2}+23\text{y}^{2}+48+36\text{x}-50\text{y}-61=0$
This is the required equation of the hyperbola.

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