# Find the equation for the ellipse that satisfies the given conditions: Length of minor axis 16, foci (0, ± 6)

**Solution:**

Length of major axis 26,

Foci (0, ±6)

Since the foci are on the y-axis, the major axis is along the y-axis

Therefore,

the equation of the ellipse will be of the form x^{2}/b^{2} + y^{2}/a^{2} = 1 where a is the semi major axis

Accordingly,

2b = 16 ⇒ b = 8 and c = 6

It is known that a^{2} = b^{2} + c^{2}

Hence,

*⇒ a*^{2} = 8^{2} + 6^{2}

*⇒ a*^{2} = 64 + 36

*⇒ a*^{2} = 100

⇒ a = √100 = 10

Thus, the equation of the ellipse is x^{2}/8^{2} + y^{2}/(10)^{2} = 1 or x^{2}/64 + y^{2}/100 = 1

NCERT Solutions Class 11 Maths Chapter 11 Exercise 11.3 Question 16

## Find the equation for the ellipse that satisfies the given conditions: Length of minor axis 16, foci (0, ± 6).

**Summary:**

The equation of the ellipse is x^{2}/64 + y^{2}/100 = 1 while the Length of major axis 16, foci (0, ± 6)