The value of lambda for which the curve (7x+5 | vedclass

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(B) The given equation is $(7x+5)^{2}+(7y+3)^{2}=\lambda^{2}(4x+3y-24)^{2}$.This is of the form $SP^{2} = \lambda^{2} PM^{2}$, where $S$ is the focus,

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

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(B) The given equation is $(7x+5)^{2}+(7y+3)^{2}=\lambda^{2}(4x+3y-24)^{2}$.This is of the form $SP^{2} = \lambda^{2} PM^{2}$, where $S$ is the focus, $P$ is a point on the curve, and $PM$ is the perpendicular distance from $P$ to the directrix.For the curve to be a parabola, the eccentricity $e$ must be equal to $1$.Here, the distance from the point $(x, y)$ to the focus $S$ is $\sqrt{(x - x_s)^2 + (y - y_s)^2}$.Rewriting the equation: $\frac{(7x+5)^{2}+(7y+3)^{2}}{49} = \lambda^{2} \frac{(4x+3y-24)^{2}}{49}$.$\left(\sqrt{(x + 5/7)^2 + (y + 3/7)^2}\right)^2 = \lambda^{2} \left(\frac{4x+3y-24}{7}\right)^2$.Comparing this with the definition of a conic $SP = e PM$, we have $e = \lambda \cdot \frac{\sqrt{4^2+3^2}}{7} = \lambda \cdot \frac{5}{7}$.For a parabola, $e = 1$, so $\lambda \cdot \frac{5}{7} = 1$.Thus, $\lambda = \pm \frac{7}{5}$.

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

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