JEE Main · 2021 · 20 Jul 2021 · Shift 1 · Parabola

Let

y = m x + c, m > 0

be the focal chord of

y^{2} = - 64 x

, which is tangent to

(x + 10)^{2} + y^{2} = 4

. Then, the value of

42 (m + c)

is equal to ____

Answer: 34

Solution

The parabola

y^{2} = - 64 x

has

4 a = 64

, so it opens leftwards with focus

(- 16, 0)

, and a focal chord

y = m x + c

must pass through that focus:

0 = - 16 m + c

, so

c = 16 m

. The chord

m x - y + 16 m = 0

touches the circle of centre

(- 10, 0)

and radius

2

when the distance from the centre equals the radius:

\frac{∣ - 10 m + 16 m ∣}{m ^{2} + 1} = 2 \Rightarrow 36 m^{2} = 4 (m^{2} + 1) \Rightarrow m^{2} = \frac{1}{8}

Taking

m > 0

gives

m = \frac{1}{2 2}

and

c = 16 m = 42

, so

42 (m + c) = 42 (\frac{1}{2 2} + 42) = 2 + 32 = 34

and the required value is

34

JEE Main · 2021 · 20 Jul 2021 · Shift 1 · Integer answer

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