Test Index

KCET 2016 Math Solved Paper

© examsnet.com

Question : 21 of 60

Marks: +1, -0

The length of latus rectum of the parabola

4y^2 + 3x + 3y + 1 = 0

is

$\frac{4}{3}$
7
12
$\frac{3}{4}$

Solution:

Given parabola

4y^{2}+3x+3y+1=0

\Rightarrow y^{2}+\;\frac{3}{4}y+\;\frac{1}{4}=\;\frac{-3}{4}x

\Rightarrow \left(y+\;\frac{3}{8}\right)^{2}-\left(\;\frac{3}{8}\right)^{2}+\;\frac{1}{4}=\;\frac{-3}{4}x

\Rightarrow \left(y+\;\frac{3}{8}\right)^{2}+\;\frac{7}{64}=\;\frac{-3}{4}x

\Rightarrow \left(y+\;\frac{3}{8}\right)^{2}=-\;\frac{3}{4}x-\;\frac{7}{64}

\Rightarrow \left(y+\;\frac{3}{8}\right)^{2}=\;\frac{-3}{4}\left(x+\;\frac{7}{48}\right)

\Rightarrow \left(y+\;\frac{3}{8}\right)^{2}=\;\frac{-3}{4}\left(x+\;\frac{7}{48}\right)

→ (1)

Comparing Eq. (1) with standard equation of parabola,

y^2=4aX

Therefore, length of latus rectum is

\frac{3}{4}

© examsnet.com

Go to Question:

Prev Question

Next Question