Ask Questions, Get Answers

Want to ask us a question? Click here
Browse Questions
0 votes

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

Can you answer this question?

1 Answer

0 votes
  • 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 : $5x-18y-28=0$
answered Jun 20, 2013 by sreemathi.v

Related questions

Ask Question
student study plans
JEE MAIN, CBSE, NEET Mobile and Tablet App
The ultimate mobile app to help you crack your examinations
Get the Android App