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Home  >>  CBSE XII  >>  Math  >>  Application of Derivatives
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Find the equations of the tangent and normal to the hyperbola \(\large {\frac{x^2}{a^2}} -\large { \frac{y^2}{b^2}} =\normalsize 1\) at the \( (x_0, \: y_0)\).

$\begin{array}{1 1} (A)\;\frac{xx_1}{a^2}-\frac{yy_1}{b^2}=1;\frac{y-y_1}{a^2y_1}+\frac{x-x_1}{b^2x_1} = 0 \\ (B)\;\large\frac{xx_1}{a^2}+\frac{yy_1}{b^2}=1;\large\frac{y-y_1}{a^2y_1}+\frac{x-x_1}{b^2x_1}=0 \\ (C)\;\large\frac{xx_1}{a^2}-\frac{yy_1}{b^2}=1;\large\frac{y-y_1}{a^2y_1}-\frac{x-x_1}{b^2x_1}=0 \\ (D)\;\large\frac{xx_1}{a^2}-\frac{yy_1}{b^2}=-1;\large\frac{y-y_1}{a^2y_1}+\frac{x-x_1}{b^2x_1}=0 \end{array} $

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1 Answer

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Toolbox:
  • Equation of the tangent at $(x_1,y_1)$ where slope is $m$ is given by $y-y_1=m(x-x_1)$
  • Equation of the normal at $(x_1,y_1)$ where slope is $m$ is given by $y-y_1=\large\frac{-1}{m}$$(x-x_1)$
Step 1:
Equation of the curve is $\large\frac{x^2}{a^2}-\frac{y^2}{b^2}$$=1$
Differentiating w.r.t $x$ we have,
$\large\frac{2x}{a^2}-\frac{2y}{b^2}\frac{dy}{dx}$$=0$
$\Rightarrow \large\frac{dy}{dx}=\frac{b^2}{a^2}\big(\large\frac{x}{y}\big)$
Slope of the tangent at$(x_1,y_1)$ is
$\large\frac{dy}{dx}_{(x_1,y_1)}=\frac{b^2}{a^2}\big(\large\frac{x_1}{y_1}\big)$
Since $(x_1,y_1)$ lies on the curve $\large\frac{x^2}{a^2}-\frac{y^2}{b^2}$$=1$,we have
$\large\frac{x_1^2}{a^2}-\frac{y_1^2}{b^2}$$=1$
Step 2:
Therefore equation of the tangent is $y-y_1=\large\frac{b^2}{a^2}\big(\large\frac{x_1}{y_1}\big)$$(x-x_1)$
On simplifying we get,
$a^2yy_1-a^2y_1^2=b^2xx_1-b^2x_1^2$
$\Rightarrow \large\frac{yy_1}{b^2}-\frac{y_1^2}{b^2}=\frac{xx_1}{a^2}-\frac{x_1^2}{a^2}$
$\Rightarrow \large\frac{xx_1}{a^2}-\frac{yy_1}{b^2}=\frac{x_1^2}{a^2}-\frac{y_1^2}{b^2}$
But $\large\frac{x_1^2}{a^2}-\frac{y_1^2}{b^2}$$=1$
$\Rightarrow \large\frac{xx_1}{a^2}-\frac{yy_1}{b^2}$$=1$
Step 3:
Now slope of the normal is $-\large\frac{1}{\Large\frac{b^2x_1}{a^2y_1}}$
$\Rightarrow -\large\frac{a^2y_1}{b^2x_1}$
Equation of th normal is $y-y_1=\large\frac{-a^2y_1}{b^2x_1}$$(x-x_1)$
On simplifying we get,
$b^2x_1(y-y_1)=-a^2y_1(x-x_1)$
$\Rightarrow \large\frac{y-y_1}{a^2y_1}+\frac{x-x_1}{b^2x_1}$$=0$
answered Jul 12, 2013 by sreemathi.v
 

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