Coordinate Geometry: JEE Main 2021 Mathematics question

JEE Main 2021 · 20 Jul, Morning shift · Q45

JEE Main 202120 Jul, Morning shiftMathematicsMediumCoordinate Geometry478× since 2002

Let y = mx + c, m > 0 be the focal chord of y² =

-

64x, which is tangent to (x + 10)² + y² = 4. Then, the value of 4

\sqrt 2

(m + c) is equal to _____________.

Your answerType your answer. Integers and decimals are accepted.

JEE Main 2021 · 20 Jul, Morning shift · Q45

JEE Main 202120 Jul, Morning shiftMathematicsMediumCoordinate Geometry478× since 2002

Let y = mx + c, m > 0 be the focal chord of y² =

-

64x, which is tangent to (x + 10)² + y² = 4. Then, the value of 4

\sqrt 2

(m + c) is equal to _____________.

Your answerType your answer. Integers and decimals are accepted.

More on Coordinate Geometry