The equation of the common tangent to the curves y ^ 2 = 4x and x ^ 2 + 32y = 0 is x + by + c = 0 . The value of | ^ – 1 ( 1 ) + ^ – 1 ( sinb ) + ^ – 1 ( sinc ) | is equal to

Correct answer: 1 y = mx + 1 m is tangent to y ^ 2 = 4x Now, x ^ 2 + 32 (mx + 1 m ) = 0 ⇒ x ^ 2 + 3 2 mx + (3 2)/(m) = 0 ⇒ (3 2 m) ^ 2 - 4 × 3 2 m = 0 ⇒ m = (1)/(2) Equation of the tangent is x – 2y + 4 = 0 ⇒ | ^ - 1 ( 1 ) + ^ - 1 ( ( - 2 ) ) + ^ - 1 ( 4 ) | = | 1 + 2 – π + π – 4 | = 1

NTA Abhyas 2020

Exam: NTA Abhyas
Year: 2020
Subject: Mathematics
Chapter: Parabola

NTA Abhyas 2020 · Mathematics · Parabola

The equation of the common tangent to the curves y ^ 2 = 4x and x ^...

The equation of the common tangent to the curves

{y}^{2} = 4x

and

{x}^{2} + 32y = 0

x + by + c = 0

. The value of

| \sin^{– 1} ( \sin 1 ) + \sin^{– 1} ( sinb ) + \sin^{– 1} ( sinc ) |

is equal to

Correct answer

Step-by-step solution

\text{y} = \text{mx} + \dfrac{1}{\text{m}}

is tangent to

{y}^{2} = 4x

Now,

{\text{x}}^{2} + \text{32} \left(\text{mx} + \dfrac{1}{\text{m}}\right) = 0 ⇒ {\text{x}}^{2} + 3 2 \text{mx} + \dfrac{3 2}{m} = 0

⇒ {\left(3 2 \text{m}\right)}^{2} - \dfrac{4 \times 3 2}{\text{m}} = 0 ⇒ \text{m} = \dfrac{1}{2}

Equation of the tangent is

x – 2y + 4 = 0

⇒ | \sin^{- 1} ( \sin 1 ) + \sin^{- 1} ( \sin ( - 2 ) ) + \sin^{- 1} ( \sin 4 ) |

= | 1 + 2 – \pi + \pi – 4 | = 1

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