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Home  >>  JEEMAIN and AIPMT  >>  Mathematics  >>  Class11  >>  Coordinate Geometry
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If the line $y=mx+c$ touches the hyperbola $\large\frac{x^2}{a^2}+\frac{y^2}{b^2}$$=1$ then find the value of c .

$\begin{array}{1 1}(a)\;\pm \sqrt{a^2m^2-b^2}\\(b)\;\pm \sqrt{a^2m^2+b^2}\\(c)\;\pm \sqrt{b^2m^2-a^2}\\(d)\;\text{None of these}\end{array}$

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Toolbox:
  • For equal roots of the quadratic equation $Ax^2+Bx+c=0$, Discriminant $B^2-AC=0$
Given Equation of the hyperbola is $\large\frac{x^2}{a^2}-\large\frac{y^2}{b^2}$$=1$...(i)
Also given that $y=mx+c$......(ii) is tangent to the hyperbola
Solving (i) and (ii) we get
$\large\frac{x^2}{a^2}-\frac{(mx+c)^2}{b^2}$$=1$
$\Rightarrow\:b^2x^2-a^2(mx+c)^2=a^2b^2$
$\Rightarrow\:(a^2m^2-b^2)x^2+2mca^2x+a^2(b^2+c^2)=0$
If the line touches the hyperbola,then the roots of this quadratic eqn. are equal.
$\Rightarrow\: $ The discriminant is $0.$
$B^2=4AC$
$\Rightarrow\:4m^2c^2a^4-4(a^2m^2-b^2)a^2(b^2+c^2)=0$
$\Rightarrow\:c^2=a^2m^2-b^2$
$\Rightarrow\:c=\pm \sqrt{a^2m^2-b^2}$
Hence (a) is the correct answer.
answered Feb 10, 2014 by sreemathi.v
edited Mar 27, 2014 by rvidyagovindarajan_1
 

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