Question

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

• A. $\left\lbrack \frac{1}{2},\frac{3}{2} \right\rbrack$
• B. $\left( \frac{1}{2},\frac{3}{2} \right)$
• C. (1, 3)
• D. $\left( \frac{–3}{2},\frac{–1}{2} \right)$

Answer 1

Answer: (B)

$\left( \frac{1}{2},\frac{3}{2} \right)$

Answer 2

The parabola is (x – 1)2 = 4$\left( y–\frac{1}{2} \right)$and origin lies outside the parabolic region; (0, 0)

makes x2 – 2x – 4y + 3 positive.

\ (2a, a) should make x2 – 2x – 4y + 3 negative

i.e., 4a2 – 8a + 3 < 0 i.e., (2a – 1) (2a – 3) < 0

\ a belongs to the open interval$\left( \frac{1}{2},\frac{3}{2} \right)$