If the length of the latus rectum of the ellipse $x^{2}+4 y^{2}+2 x+8 y-\lambda=0$ is 4 , and $l$ is

Question

If the length of the latus rectum of the ellipse $x^{2}+4 y^{2}+2 x+8 y-\lambda=0$ is 4 , and $l$ is the length of its major axis, then $\lambda+l$ is equal to

Professor PC Thomas Classes · Answer

(75) : Given, ellipse is\ $x^{2}+4 y^{2}+2 x+8 y-\lambda=0$\ $\Rightarrow \quad(x+1)^{2}+4(y+1)^{2}=\lambda+5$\ $\Rightarrow \quad \frac{(x+1)^{2}}{\lambda+5}+\frac{(y+1)^{2}}{\frac{\lambda+5}{4}}=1$\ Given, length of major axis $l=2 \sqrt{\lambda+5}$\ Given, length of latus rectum $=4$\ $\Rightarrow \frac{2 	imes\left(\frac{\lambda+5}{4}ight)}{\sqrt{\lambda+5}}=4 \Rightarrow 4 \sqrt{\lambda+5}=\frac{\lambda+5}{2}$\ $\Rightarrow 8=\sqrt{\lambda+5} \Rightarrow \lambda=59$\ So, $l=2 \sqrt{59+5}=16$\ $	herefore \quad \lambda+l=59+16=75$\