# Find the points on the curve at which the tangents are x²/9 + y²/16 = 1

(i) parallel to x-axis (ii) parallel to y-axis

**Solution:**

The slope of a line is nothing but the change in y coordinate with respect to the change in x coordinate of that line.

The equation of the given curve is

x^{2}/9 + y^{2}/16 = 1

On differentiating both sides with respect to x, we have:

2x / 9 + 2y / 16 dy/dx = 0

dy/dx = - 16x / 9y

(i) The tangent is parallel to the x-axis if the slope of the tangent is

- 16x/9y = 0,

which is possible if x = 0

Thus,

x^{2}/9 + y^{2}/16 = 1 for x = 0

Therefore,

⇒ y^{2}/16 = 1

⇒ y^{2} = 16

⇒ y = ± 4

Hence, the points are (0, 4) and (0, - 4).

(ii) The tangent is parallel to the y-axis if the slope of the normal is 0, which gives

- 1/(- 16x / 9y) = 0

9y / 16x = 0

y = 0

Thus,

x^{2}/9 + y^{2}/16 = 1 for y = 0

Therefore,

x^{2}/9 = 1

⇒ x^{2} = 9

⇒ x = ± 3

Hence, the points are (3, 0) and (- 3, 0)

NCERT Solutions Class 12 Maths - Chapter 6 Exercise 6.3 Question 13

## Find the points on the curve at which the tangents are x²/9 + y²/16 = 1 (i) parallel to x-axis (ii) parallel to y-axis

**Summary:**

The points are (0, 4) and (0, - 4) on the curve at which the tangents are x^{2}/9 + y^{2}/16 = 1 parallel to x-axis and (3, 0) and (- 3, 0) on the curve at which the tangents are x^{2}/9 + y^{2}/16 = 1 parallel to y-axis