Question

Set of values of ‘h’ for which the number of distinct common normals of

(x – 2)² = 4(y – 3) and x² + y² – 2x – hy – c = 0 (c > 0) is 3, is

• A. (2, ¥)
• B. (4, ¥)
• C. (2, 4)
• D. (10, ¥)

Answer 1

Answer: (D)

(10, ¥)

Answer 2

The equation of any normal

(x – 2)² = 4(y – 3) is x – 2 = m (y – 3) – 2m – m³

If it passes through (1, h/2), then

1 – 2 = m $\left( \frac{h}{2} - 3 \right)$ – 2m – m³ Ž 2m³ + m

(10 – h) – 2 = 0

This equation will give three distinct values of m, if Ģ (m)

= 0 has two distinct roots,

where ƒ(m) = 2m³ + m (10 – h) – 2

Now ƒ¢(m) = 6m² + (10 – h)

Ģ (m) = 0

Ž m ± $\sqrt{\frac{h - 10}{6}}$

The values of m are real and distinct if h > 10 i.e. h Ī (10, ).

Hence (4) is correct answer.