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Home  >>  JEEMAIN and NEET  >>  Mathematics  >>  Class11  >>  Coordinate Geometry

Find the value of $\lambda$ such that $\Delta =0$ for the equation $x^2+y^2-3x+2y+\lambda=0$

$\begin{array}{1 1}(a)\;\large\frac{13}{5}&(b)\;\large\frac{13}{4}\\(c)\;\large\frac{11}{4}&(d)\;\large\frac{13}{3}\end{array}$

1 Answer

$\Delta=abc+2fgh-af^2-bg^2-ch^2=0$
$1\times 1\times \lambda+2\times 0-1\times 1^2-1\times (-\large\frac{3}{2})^2$$-0=0$
$\lambda -1-1\times \large\frac{9}{4}$$=0$
$\lambda=1+\large\frac{9}{4}$
$\lambda=\large\frac{13}{4}$
Hence (b) is the correct option.
answered Feb 5, 2014 by sreemathi.v
 
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