If $x = t^{2}$ and $y = 2 t$ , then equation of the normal at $t = 1$ is

x + y - 3 = 0

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

x + y - 1 = 0

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

x + y + 1 = 0

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

x + y + 3 = 0

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

Open in App

Solution

The correct option is A $x + y - 3 = 0$
$x = t^{2}$ , $y = 2 t$
$y^{2} = 4 t^{2} = 4 x$ $t = 1$
$y^{2} = 4 x$ $x = 1, y = 2$
$2 y \frac{d y}{d x} = 4 \to \frac{d y}{d x} = \frac{2 y}{y}$
$y = 2$
$\frac{d y}{d x} = 1$
$- \frac{d x}{d y} = - 1$
equation of normal
$y - 2 = - 1 (x - 1)$
$y - 2 + x - 1 = 0$
$x + y - 3 = 0$

Suggest Corrections

Similar questions

x = t^{2}

x = t^{2}

y = 2 x^{2} + 3 sin x

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

x + 2 y = 3

Equation of the line passing through (1, 2) and parallel to the line $y = 3 x - 1$ is .

Join BYJU'S Learning Program