Question

The length of the latusrectum of the parabola is 2y^2+3y+4x-2=0

Answer 1

SOLUTION

TO DETERMINE

The length of the latus rectum of the parabola

$\sf{2 {y}^{2} + 3y + 4x - 2 = 0}$

EVALUATION

Here the given equation of parabola is

$\sf{2 {y}^{2} + 3y + 4x - 2 = 0}$

Which can be rewritten as below

$\sf{2 {y}^{2} + 3y + 4x - 2 = 0}$

$\displaystyle \implies \: \sf{ {y}^{2} + \frac{3y}{2} + 2x - 1 = 0}$

$\displaystyle \implies \sf{ {y}^{2} + 2.y.\frac{3}{4} + { \bigg( \frac{3}{4} \bigg)}^{2} -{ \bigg( \frac{3}{4} \bigg)}^{2} + 2x - 1 = 0}$

$\displaystyle \implies \sf{ { \bigg( y + \frac{3}{4} \bigg)}^{2} - \frac{9}{16} + 2x - 1 = 0}$

$\displaystyle \implies \sf{ { \bigg( y + \frac{3}{4} \bigg)}^{2} + 2x - \frac{25}{16} = 0}$

$\displaystyle \implies \sf{ { \bigg( y + \frac{3}{4} \bigg)}^{2} = - 2x + \frac{25}{16} }$

$\displaystyle \implies \sf{ { \bigg( y + \frac{3}{4} \bigg)}^{2} = - 2 \bigg(x - \frac{25}{32} \bigg)}$

Comparing with

$\sf{ {(y - \beta )}^{2} = - 4a(x - \alpha )}$

We get

$\sf{4a = 2}$

Hence the length of the latus rectum = 2 unit

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