Question

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

$\pm \frac{7}{5}$

Explanation for the correct option:

Given, ${\left(7x+5\right)}^2+{\left(7y+3\right)}^2={\lambda }^2{(4x+3y-24)}^2$

Rewrite the given equation as follows:

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

We know that for the equation $a{x}^2+2hxy+b{y}^2+2gx+2fy+c=0$ the condition of the parabola is $ab=h^2$.

Therefore,

$\begin{array}{cc}\Rightarrow & (49-16{\lambda }^2)(49-9{\lambda }^2)={(-12{\lambda }^2)}^2\\ \Rightarrow & 2401-441{\lambda }^2-78{\lambda }^2+144{\lambda }^4=144{\lambda }^4\\ \Rightarrow & 2401-1225{\lambda }^2=0\\ \Rightarrow & {\lambda }^2=\frac{2401}{1225}\\ \Rightarrow & {\lambda }^2=\frac{{\left(49\right)}^2}{{\left(35\right)}^2}\\ \Rightarrow & \lambda =\pm \frac{7}{5}\end{array}$

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

Hence, option $B$ is the correct answer.