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Find the equation of the tangent and normal at $(3 , 4 )$ to the rectangular hyperbola $xy=12$

1 Answer

  • 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}$$+F=0$
  • The normal is the perpendicular to this line that passes through $(x_1,y_1)$.
  • Any line perpendicular to $ax+by+c=0$ is of the form $bx-ay+k=0$.This can be used to find the equation of the normal.
Step 1:
Equation of the rectangular hyperbola is $xy=12$.
Equation of the tangent to $xy=12$ at $(x_1,y_1)$ is $\large\frac{1}{2}$$(xy_1+yx_1)$$=12$
Here $(x_1,y_1)=(3,4)$
Therefore the tangent is $4x+3y=24$
Step 2:
The normal is $\perp$ to the tangent and the equation of the form $3x-4y=k$.It passes through $(3,4)$
Therefore $9-16=k$
$\Rightarrow k=-7$
The normal is $3x-4y=-7$
$\Rightarrow 3x-4y+7=0$
Step 3:
Equation of tangent : $4x+3y=24$
Equation of normal : $3x-4y+7=0$
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