If the length of the latus rectum of the ellipse x^2+4y^2+2x+8y-=0 is 4, and l is the leng…

Question

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

Answer: 75

Accepted Answer

$(x+1)^2+4(y+1)^2=\lambda+5$, i.e. $\dfrac{(x+1)^2}{\lambda+5}+\dfrac{(y+1)^2}{(\lambda+5)/4}=1$, with $a^2=\lambda+5$ (major, along $x$).

$$\text{LR}=\frac{2b^2}a=\frac{(\lambda+5)/2}{\sqrt{\lambda+5}}=\frac{\sqrt{\lambda+5}}2=4\ \Longrightarrow\ \lambda+5=64\ \Longrightarrow\ \lambda=59.$$

$$l=2a=2\sqrt{64}=16,\qquad \lambda+l=59+16=75.$$

Solutions · 1