You visited us 1 times! Enjoying our articles? Unlock Full Access!

Fundamental Laws of Logarithms

If x, y is ...

Question

If $(x, y)$ is a solution of $x^{2} + y^{2} = 160$ and $y = 3 x$ , then $y^{2}$ is equal to

12

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

16

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

120

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

144

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

Open in App

Solution

The correct option is C $144$
Let $x^{2} + y^{2} = 160$ ---(1)
We have $y = 3 x$
Substitute the value of $y$ in the above equation, we get
$x^{2} + (3 x)^{2} = 160$
$∴ 10 x^{2} = 160$
$∴ x^{2} = 16$
Plug the value of $x^{2}$ in equation (1), we get:
$16 + y^{2} = 160$
$∴ y^{2} = 160 - 16$
$∴ y^{2} = 144$

Suggest Corrections

Similar questions

{(x + y)}^{2} = 144

x^{2} - y^{2} = 24

x

x

y

2^{2 x - y} = 32

2^{x + y} = 16

x^{2} + y^{2}

sin 21^{\circ} = \frac{x}{y}

sec 21^{\circ} - sin 69^{\circ}

If A = { $(x, y)$ : $x^{2} + y^{2} = 25$ } And B = {(x,y): $x^{2} + 9 y^{2} = 144$ }, then $A \cap B$ contains

x^{y} = y^{x},

(\frac{x}{y})^{\frac{x}{y}}

\frac{x}{y} = \frac{6}{5}

\frac{x^{2} + y^{2}}{x^{2} - y^{2}}

Join BYJU'S Learning Program