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Find the equation of the tangent and normal at $(-2 , \large\frac{1}{4})$ to the rectangular hyperbola $ 2xy-2x-8y-1=0 $

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Toolbox:
  • The equation of the tangent at $(x_1,y_1)$ to a conic $Ax^2+Bxy+Cy^2+Dx+Ey+F=0$ is given by $Axx_1+\large\frac{B}{2}$$(xy_1+x_1y)+Cyy_1+D\large\frac{(x+x_1)}{2}+$$E\large\frac{(y+y_1)}{2}$$+F=0$
  • The normal is the perpendicular to this line that passes through $(x_1,y_1)$.
  • Any line perpendicular to $ax+by+c=0$ is of the form $bx-ay+k=0$.This can be used to find the equation of the normal.
Step 1:
Equation of the tangent to $2xy-2x-8y-1=0$ at $(x_1,y_1)$
$\large\frac{1}{2}$$\times 2(xy_1+yx_1)-2\large\frac{(x+x_1)}{2}$$-8\large\frac{(y+y_1)}{2}$$-1=0$
Here $(x_1,y_1)=(-2,\large\frac{1}{4})$
The tangent is $(\large\frac{x}{4}-$$2y)-(x-2)-4(y+\large\frac{1}{4})$$-1=0$
$x-8y-4x+8-16y-4-4=0$
$-3x-24y=0$
$3x+24y=0$
$3(x+8y)=0$
$\Rightarrow x+8y=0$
Step 2:
The normal is $\perp$ to the tangent.Its equation is of the form $8x-y+k=0$
It passes through $(-2,\large\frac{1}{4})$.
Therefore $8(-2)-\large\frac{1}{4}$$+k=0$
$k=16+\large\frac{1}{4}=\large\frac{65}{4}$
The equation of the normal is $8x-y+\large\frac{15}{4}$$=0$
$\Rightarrow 32x-4y+65=0$
Step 3:
Equation of tangent $x+8y=0$
Equation of normal $32x-4y+65=0$
answered Jun 24, 2013 by sreemathi.v
 

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