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Home  >>  JEEMAIN and AIPMT  >>  Mathematics  >>  Class11  >>  Coordinate Geometry
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Find the equation and length of common tangent to hyperbola $\large\frac{x^2}{p^2}-\frac{y^2}{q^2}$$=1$ and $\large\frac{y^2}{p^2}-\frac{x^2}{q^2}$$=1$

$\begin{array}{1 1}(a)\;x\mp y=\pm\sqrt{p^2-q^2},\sqrt 2\large\frac{p^2+q^2}{p^2-q^2}\\(b)\;x\mp y=\pm\sqrt{p^2+q^2},\sqrt 2\large\frac{p^2-q^2}{p^2+q^2}\\(c)\;x\pm y=\mp\sqrt{p^2-q^2},\sqrt 2\large\frac{p+q}{p^2-q^2}\\(d)\;\text{None of these}\end{array}$

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Tangent at $(p\sec\theta,q\tan\phi)$ for first hyerbola
$\large\frac{x\sec\phi}{p}-\frac{y\tan \phi}{q}$$=1$------(1)
Similarly at $(q\tan\theta,p\sec\theta)$ for second hyperbola
$\large\frac{y\sec\theta}{p}-\frac{x\tan \theta}{q}$$=1$------(2)
If (1) and (2) are common tangent then they should be identical hence
$\large\frac{\sec\theta}{p}=-\frac{\tan\phi}{q}$
$\sec\theta=-\large\frac{p\tan\phi}{q}$------(3)
$-\large\frac{\tan\theta}{q}=\frac{\sec\theta}{p}$
$\tan\theta=-\large\frac{\sec\phi}{p}$
$\tan\theta=-\large\frac{-q}{p}$$\sec\phi$------(4)
$\sec^2\theta-\tan^2\theta=1$
$\large\frac{p^2}{q^2}$$\tan^2\phi-\large\frac{q^2}{p^2}$$\sec^2\phi=1$
$\large\frac{p^2}{q^2}$$\tan^2\phi-\large\frac{q^2}{p^2}$$(1+\tan^2\phi)=1$
$\tan^2\phi=\large\frac{q^2}{p^2-q^2}$
$\sec^2\phi=\large\frac{p^2}{p^2-q^2}$
Hence point of contact are $\big(\pm\large\frac{p^2}{\sqrt{p^2-q^2}},\pm\large\frac{q^2}{\sqrt{p^2-q^2}}\big),\big(\pm\large\frac{q^2}{\sqrt{p^2-q^2}},\pm\large\frac{p^2}{\sqrt{p^2-q^2}}\big)$
Length of common tangent is $\large\frac{\sqrt 2(p^2+q^2)}{p^2-q^2}$
Equation of common tangent is
$\pm \large\frac{x}{\sqrt{p^2-q^2}}\mp\frac{y}{\sqrt{p^2-q^2}}$$=1$
$x\mp y=\pm \sqrt{p^2-q^2}$
Hence (a) is the correct answer.
answered Feb 10, 2014 by sreemathi.v
 

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