The angle of intersection between the curves $y^{2} = 16 x a n d 2 x^{2} + y^{2} = 4$ is -

0^{\circ}

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

30^{\circ}

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

45^{\circ}

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

90^{\circ}

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

Open in App

Solution

The correct option is D $90^{\circ}$
Let $c_{1} : y^{2} = 16 x$ and $c_{2} : 2 x^{2} + y^{2} = 4$
Differentiating w.r.t $x$
$c_{1} : 2 y \frac{d y}{d x} = 16$ and $c_{2} : 4 x + 2 y \frac{d y}{d x} = 0$
$m_{1} = {(\frac{d y}{d x})}_{c_{1}} = \frac{8}{y}$ and $m_{2} = {(\frac{d y}{d x})}_{c_{2}} = - \frac{2 x}{y}$
$∴ m_{1} . m_{2} = - \frac{16 x}{y^{2}} = - 1$ , since the point lies on the curve $c_{1}$ and $c_{2}$
Therefore, both the curve intersect on right angle.
Hence, option 'D' is correct.

Suggest Corrections

Similar questions

y^{2} = 8 x a n d x^{2} = 4 y - 12

(2, 4)

y^{2} = 4 x

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

x^{2} + y^{2} = a^{2} \sqrt{2}

x^{2} - y^{2} = a^{2} .

S_{1} : x^{2} + y^{2} - 4 x + 6 y + 11 = 0

S_{2} : x^{2} + y^{2} - 2 x + 8 y + 13 = 0

Join BYJU'S Learning Program