Question

The value of $\lambda$ for which the curve ${(7x+5)}^2+{(7y+3)}^2={\lambda }^2{(4x+3y-24)}^2$ represents a parabola is

• A. $\pm \frac{6}{5}$
• B. $\pm \frac{7}{5}$
• C. $\pm \frac{1}{5}$
• D. $\pm \frac{2}{5}$

Answer 1

The correct option is B $\pm \frac{7}{5}$

Given curve is
${(7x+5)}^2+{(7y+3)}^2={\lambda }^2{(4x+3y-24)}^2$
$\Rightarrow 49x^2+25+70x+49y^2+9+42y$
$={\lambda }^2(16x^2+9y^2+576+24xy-144y-192x)$

$\Rightarrow (49-16{\lambda }^2)x^2+(49-9{\lambda }^2)y^2+(70+192{\lambda }^2)x+(42+144{\lambda }^2)y-24{\lambda }^{2}xy+(25-576{\lambda }^2)=0$

On comparing with
$a{x}^2+2hxy+b{y}^2+2gx+2fy+c=0$, we get
$a=49-16{\lambda }^2,b=(49-9{\lambda }^2),h=-12{\lambda }^2$

Condition for parabola is

$ab=h^2$

$\therefore (49-16{\lambda }^2)(49-9{\lambda }^2)={(-12{\lambda }^2)}^2$
$\Rightarrow 2401-441{\lambda }^2-78{\lambda }^2+144{\lambda }^4=144{\lambda }^4$
$\Rightarrow 2401-1225{\lambda }^2=0$

$\Rightarrow {\lambda }^2=\frac{2401}{1225}$ $\Rightarrow {\lambda }^2=\frac{{\left(49\right)}^2}{{\left(35\right)}^2}$

$\Rightarrow \lambda =\pm \frac{49}{35}$

$\Rightarrow \lambda =\pm \frac{7}{5}$