Question

The latus-rectum of the conic 3x² + 4y² − 6x + 8y − 5 = 0 is

(a) 3

(b) $\frac{\sqrt{3}}{2}$

(c) $\frac{2}{\sqrt{3}}$

(d) none of these

Answer 1

$$\begin{gathered} \left(\mathrm{a}\right)3 \\ 3x^2+4y^2-6x+8y-5=0 \\ \Rightarrow 3(x^2-2x)+4(y^2+2y)=5 \\ \Rightarrow 3(x^2-2x+1)+4(y^2+2y+1)=5+3+4 \\ \Rightarrow 3(x-1{)}^2+4(y+1{)}^2=12 \\ \Rightarrow \frac{(x-1{)}^2}{4}+\frac{(y+1{)}^2}{3}=1 \\ \mathrm{So},a=2\mathrm{and}b=\sqrt{3} \\ \therefore \text{Latus rectum}=\frac{2b^2}{a} \\ =2\frac{{\left[\sqrt{3}\right]}^2}{2} \\ =3 \end{gathered}$$