Question

The latus rectum of a conic section is the width of the function through the focus. The positive difference between the length of the latus rectum of $3y=x^2+4x-9$ and $x^2+4y^2-6x+16y=24$ is

• A. $\frac{1}{2}$
• B. $2$
• C. $\frac{3}{2}$
• D. $\frac{5}{2}$

Answer 1

The correct option is A $\frac{1}{2}$

$3x=x^2+4x-9$

$3y=(x+2{)}^2-13$

$3y+13=(x+2{)}^2$

$(x+2{)}^2=3(y+\frac{13}{3})\Rightarrow$ latus Rectum $=3$

The other conic is,

$(x-3{)}^2+4(y^2+4y)=24+9$

$(x-3{)}^2+4(y+2{)}^2=49$

$\frac{(x-3{)}^2}{7^2}+\frac{(y+2{)}^2}{{\left(\frac{7}{2}\right)}^2}=1$ which is an ellipse

Latus Rectum $=\frac{2b^2}{a}=\frac{2.49}{4.7}=\frac{7}{2}$

$\therefore$ Positive difference $=\frac{7}{2}-3=\frac{1}{2}$