# Find the equation of tangent to the curve y = x2 + 4x at the point whose ordinate is – 3 - Mathematics and Statistics

Sum

Find the equation of tangent to the curve y = x2 + 4x at the point whose ordinate is – 3

#### Solution

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

Differentiating w.r.t. x, we get

("d"y)/("d"x) = 2x + 4

y = – 3    ......[Given]

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

– 3 = x2 + 4x

∴ x2 + 4x + 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 ("d"y)/("d"x) = 2(– 1) + 4 = 2

∴ Equation of tangent at (– 1,– 3) is

y + 3 = 2(x + 1)

∴ y + 3 = 2x + 2

∴ 2x – y – 1 = 0

For x = – 3, y = (– 3)2 + 4(– 3) = – 3

∴ Point is (x, y) = (– 3, – 3)

Slope of tangent at (– 3, – 3) is ("d"y)/("d"x) = 2(– 3) + 4 = – 2

Equation of tangent at (– 3, – 3) is

y + 3 = – 2(x + 3)

∴ y + 3 = – 2x – 6

∴ 2x + y + 9 = 0

Is there an error in this question or solution?
Chapter 1.4: Applications of Derivatives - Q.5

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