Question

Length of the latus rectum of the parabola
25$\left[\right(x-2{)}^2+(y-3{)}^2]=(3x-4y+7{)}^{2}is$

• A. 4
• B. 2/5
• C. 1/5
• D. 2

Answer 1

The correct option is B 2/5

$25\left[\right(x-2{)}^2+(y-3{)}^2]=(3x-4y+7{)}^2$

$\Rightarrow (x-2{)}^2+(y-3{)}^2=\frac{(3x-4y+7{)}^2}{25}$

$\Rightarrow (x-2{)}^2+(y-3{)}^2=\frac{(3x-4y+7{)}^2}{5^2}$

$\Rightarrow (x-2{)}^2+(y-3{)}^2=\frac{(3x-4y+7{)}^2}{5^2+(-4{)}^2}.....\left(1\right)$

The above equation is in the form

$(x-\alpha {)}^2+(y-\beta {)}^2=\frac{(lx+my+n{)}^2}{l^2+m^2}......\left(2\right)$

This equation represents the parabola with focus $(\alpha ,\beta )$ & directrix $lx+my+n=0$

$\therefore$ Focus $\Rightarrow (\alpha ,\beta )=(2,3)$

Directrix $\Rightarrow 3x-4y-7=0$

[On comparing eqn $\left(1\right)$ & eqn $\left(2\right)$]

Latus rectum $=2\times$ [Shortest (perpendicular) distance focus from directrix]

$\Rightarrow 2\times \left|\frac{3\left(2\right)-4\left(3\right)+7}{\sqrt{3^2+4^2}}\right|$

$\Rightarrow 2\times \left|\frac{6-12+7}{5}\right|$

$\Rightarrow 2\times \frac{1}{5}=\frac{2}{5}$ $\left[B\right]$