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Question
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
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