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Maths-

General

Easy

Question

If the point (2a, a) lies inside the parabola x² – 2x – 4y + 3 = 0, then a lies in the interval -

$[\frac{1}{2}, \frac{3}{2}]$
$(\frac{1}{2}, \frac{3}{2})$
(1, 3)
$(\frac{- 3}{2}, \frac{- 1}{2})$

hint Hint:

If f(x,y) = 0 is the equation of parabola with axis parallel to x axis and point (a,b) be a point inside parabola then f(a,b) < 0.

The correct answer is: $open parentheses fraction numerator 1 over denominator 2 end fraction comma fraction numerator 3 over denominator 2 end fraction close parentheses$

If f(x,y) =x2 – 2x – 4y + 3 = 0 is the equation of parabola with axis parallel to x axis and point (a,b) be a point inside parabola then f(a,b) < 0.
Then f(2a,a) < 0
(2a)² – 2(2a) – 4(a) + 3 < 0
4a² - 8a + 3 < 0
(2a-1)(2a-3) < 0 then a lies in $open parentheses fraction numerator 1 over denominator 2 end fraction comma fraction numerator 3 over denominator 2 end fraction close parentheses$ .

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