JEE Main · 2021 · 26 Feb 2021 · Shift 2 · Differential Equations

Let the normals at all the points on a given curve pass through a fixed point (a, b). If the curve passes through

(3, - 3)

and

(4, - 22)

and given that

a - 22 b = 3

, then

(a^{2} + b^{2} + ab)

is equal to ____

Answer: 9

Solution

The normal at a point

P

of the curve passes through

(a, b)

exactly when

P - (a, b)

is perpendicular to the tangent there, so along the curve

\frac{d}{d x} ∣ P - (a, b) ∣^{2} = 2 (P - (a, b)) \cdot P^{'} = 0

The distance from

(a, b)

is therefore constant and the curve is a circle centred at

(a, b)

. Equating the squared distances of

(3, - 3)

and

(4, - 22)

from the centre,

(3 - a)^{2} + (3 + b)^{2} = (4 - a)^{2} + (22 + b)^{2}

18 - 6 a + 6 b = 24 - 8 a + 42 b \Rightarrow a + (3 - 22) b = 3

Subtracting the given relation

a - 22 b = 3

leaves

3 b = 0

, so

b = 0

and

a = 3

. Hence

a^{2} + b^{2} + ab = 9 + 0 + 0 = 9

The value of

a^{2} + b^{2} + ab

9

JEE Main · 2021 · 26 Feb 2021 · Shift 2 · Integer answer