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Find the equation of the tangent and normal to the hyperbola $9x^{2}-5y^{2}=31 $ at $(2 , -1 )$

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:
Tangent and normal at $(2,-1)$ to the hyperbolic $9x^2-5y^2=31$
The tangent at $(x_1,y_1)$ is $9xx_1-5yy_1=31$
Here $(x_1,y_1)=(2,-1)$.
Therefore the tangent is $9\times 2x-5\times (-1)y=31$
$\Rightarrow 18x+5y=31$
Step 2:
The normal is $\perp$ to the tangent and its equation is of the form $5x-18y+k=0$
It passes through $(2,-1)$.
Therefore $10+18+k=0$
$\Rightarrow k=-28$
The equation of the normal is $5x-18y-28=0$
Step 3:
Equation of tangent : $18x+5y-31=0$
Equation of normal : $5x-18y-28=0$
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