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…

Download App

Question

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

✓

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.