Question

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

1. $\pm \dfrac{6}{5}$
2. $\pm \dfrac{7}{5}$
3. $\pm \dfrac{1}{5}$
4. $\pm \dfrac{2}{5}$

Answer 1

Compare the given equation of the curve with the standard equation of the second order $a{x^2} + b{y^2} + 2hxy + 2gx + 2fy + c = 0$. Solve the condition ${h^2} = ab$ to find the condition on $\lambda$ for which the given equation of the curve is a parabola.

Complete step-by-step answer:
The equation of the second order represents a parabola. For the standard equation $a{x^2} + b{y^2} + 2hxy + 2gx + 2fy + c = 0$ to represent a parabola, ${h^2} = ab$ is the sufficient and necessary condition, where $h$ is the coefficient of $2xy$, $a$ is the coefficient of ${x^2}$ and $b$ is the coefficient of ${y^2}$.
For the given equation of the curve ${\left( {7x + 5} \right)^2} + {\left( {7y + 3} \right)^2} = {\lambda ^2}{\left( {4x + 3y - 24} \right)^2}$ ,we can simplify the equation as
$49{x^2} + 25 + 70x + 49{y^2} + 9 + 42y = {\lambda ^2}\left( {16{x^2} + 9{y^2} + 576 + 24xy - 144y - 192x} \right)$
On further simplifying by writing in the standard form, we get
${x^2}\left( {49 - 16{\lambda ^2}} \right) + {y^2}\left( {49 - 9{\lambda ^2}} \right) - 24{\lambda ^2}xy + x\left( {70 + 192{\lambda ^2}} \right) + y\left( {42 + 144{\lambda ^2}} \right) + 34 - 576{\lambda ^2} = 0$
On comparing the equation ${x^2}\left( {49 - 16{\lambda ^2}} \right) + {y^2}\left( {49 - 9{\lambda ^2}} \right) - 24{\lambda ^2}xy + x\left( {70 + 192{\lambda ^2}} \right) + y\left( {42 + 144{\lambda ^2}} \right) + 34 - 576{\lambda ^2} = 0$ with the standard equation $a{x^2} + b{y^2} + 2hxy + 2gx + 2fy + c = 0$, we get
$a = 49 - 16{\lambda ^2}$, $b = 49 - 9{\lambda ^2}$ and $h = 12{\lambda ^2}$
Substituting the values $a = 49 - 16{\lambda ^2}$, $b = 49 - 9{\lambda ^2}$ and $h = 12{\lambda ^2}$ in the equation ${h^2} = ab$ to find the condition on $\lambda$ for which the given equation of the curve is a parabola, we get
${\left( {12{\lambda ^2}} \right)^2} = \left( {49 - 16{\lambda ^2}} \right)\left( {49 - 9{\lambda ^2}} \right)$
Solving the equation to find the value of $\lambda$.
$144{\lambda ^4} = {49^2} - 16\left( {49{\lambda ^2}} \right) - 9\left( {49{\lambda ^2}} \right) + 144{\lambda ^4} \\\ 49{\lambda ^2}\left( {16 + 9} \right) = {49^2} \\\ {\lambda ^2}\left( {25} \right) = 49 \\\ {\lambda ^2} = \dfrac{{49}}{{25}} \\\ \lambda = \sqrt {\dfrac{{49}}{{25}}} \\\ \lambda = \pm \dfrac{7}{5} \\\$
Thus the condition for which the given equation of the curve ${\left( {7x + 5} \right)^2} + {\left( {7y + 3} \right)^2} = {\lambda ^2}{\left( {4x + 3y - 24} \right)^2}$ is a parabola is $\lambda = \pm \dfrac{7}{5}$.
Hence, option B is the correct answer.

Note: For the standard equation $a{x^2} + b{y^2} + 2hxy + 2gx + 2fy + c = 0$ to represent a parabola, ${h^2} = ab$ is the sufficient and necessary condition. The solution to the equation ${x^2} = {a^2}$ has two solutions, that are $a$ and $- a$.