The number of values of $c$ such that the straight line $y = 2 x + c$ touches the curve $4 x^{2} + y^{2} = 4$ is

0

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

1

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

2

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

infinite

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

Open in App

Solution

The correct option is C $2$
The point of intersection of the line and curve will be -
$4 x^{2} + (2 x + c)^{2} = 4$
$8 x^{2} + 4 c x + c^{2} - 4 = 0$
putting discriminant as zero, we get
$16 c^{2} - 4 \times 8 (c^{2} - 4) = 0$
$c^{2} - 2 (c^{2} - 4) = 0$
$c^{2} - 2 c^{2} + 8 = 0$
$c^{2} = 8$
$c = \pm 2 \sqrt{2}$
Hence, $2$ values

Suggest Corrections

Similar questions

c

y = 4 x + c

\frac{x^{2}}{4} + y^{2} = 1

y = 2 sin x

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

3 x + 4 y = c

\frac{x^{4}}{2} = x + y

The number of solutions of ${log}_{3} (x - 3) = 4 - x$ is

x

{log}_{2} | 4 - 5 x | > 2

Join BYJU'S Learning Program