# The Latus-rectum of the Conic 3x2 + 4y2 − 6x + 8y − 5 = 0 is - Mathematics

MCQ
Sum

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

#### Options

• 3

• $\frac{\sqrt{3}}{2}$

• $\frac{2}{\sqrt{3}}$

• none of these

#### Solution

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

Concept: Introduction of Ellipse
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#### APPEARS IN

RD Sharma Class 11 Mathematics Textbook
Chapter 26 Ellipse
Q 9 | Page 28
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