Question

Find the equation of tangent and normal to the following curve.

y = x² + 4x at the point whose ordinate is -3.

Answer 1

Equation of the curve is y = x² + 4x     ....(i)

Differentiating w.r.t. x, we get

`"dy"/"dx" = 2"x"+ 4`

y = −3       ...[Given]

Putting the value of y in (i), we get

- 3 = x² + 4x

∴ x² + 4x + 3 = 0

∴ (x + 1)(x + 3) = 0

∴ x = −1 or x = −3

For x = −1, y = (−1)² + 4(−1) = −3

∴ Point is (x, y) = (−1, −3)

Slope of tangent at (–1, –3) is `"dy"/"dx"` = 2(−1) + 4 = 2

Equation of tangent at (−1, −3) is

y + 3 = 2(x + 1)

∴ y + 3 = 2x + 2

∴ 2x − y − 1 = 0

Slope of normal at (–1, –3) is `(-1)/("dy"/"dx") = (-1)/2`

Equation of normal at (–1, –3) is

y + 3 = `(-1)/2` (x + 1)

∴ 2y + 6 = −x − 1

∴ x + 2y + 7 = 0

For x = −3, y = ( −3)² + 4(−3) = −3

∴ Point is (x, y) = (−3, −3)

Slope of tangent at (–3, –3) = 2(−3) + 4 = −2

Equation of tangent at (−3, −3) is

y + 3 = −2(x + 3)

∴ y + 3 = −2x − 6

∴ 2x + y + 9 = 0

Slope of normal at (– 3, – 3) is `(-1)/("dy"/"dx") = 1/2`

Equation of normal at (- 3, - 3) is

y + 3 = `1/2`(x + 3)

∴ 2y + 6 = x + 3

∴ x − 2y − 3 = 0