The equation of tangent to the curve $y = 3 x^{2} - x + 1$ at $P (1, 3)$ is ____

5 x - y = 2

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

x + 5 y = 16

No worries! We‘ve got your back. Try BYJU‘S free classes today!

5 x - y + 2 = 0

No worries! We‘ve got your back. Try BYJU‘S free classes today!

5 x = y

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is A $5 x - y = 2$
Given curve is $y = 3 x^{2} - x + 1$
$\frac{d y}{d x} = 6 x - 1$ , differentiate the curve
So slope of tangent at $(1, 3)$ is, $m = \frac{d y}{d x} {∣ ∣ ∣}_{(1, 3)} = 6 (1) - 1 = 5$
Hence equation of tangent passes through $(1, 3)$ is given by
$y - 3 = 5 (x - 1)$
$\Rightarrow y - 3 = 5 x - 5$
$\Rightarrow 5 x - y = 2$

Suggest Corrections

Similar questions

y (1 + x^{2}) = 2 - x

x

y^{3} - x^{2} y + 5 y - 2 x = 0

x^{4} - x^{3} y^{2} + 5 x + 2 y = 0

θ

y = 2 t^{2} + t, x = t^{2}

(1, 3)

y = x^{2} - 5 x + 6

(2, 0)

(3, 0)

Join BYJU'S Learning Program