Q

# Find points on the curve x ^ 2/ 9+ y ^ 2 / 16 = 1 at which the tangents are parallel to y-axis

13) Find points on the curve  $\frac{x^2}{9} + \frac{y^2}{16} = 1$ at which the tangents are  (b) parallel to y-axis

Views

Parallel to y-axis means the slope of the tangent is  $\infty$ , means the slope of normal is 0
We know that slope of the tangent at a given point on the given curve is given by  $\frac{dy}{dx}$
Given the equation of the curve is
$\frac{x^2 }{9} + \frac{y^2 }{16} = 1 \Rightarrow 9y^2 = 144(1-16x^2)$
$18y\frac{dy}{dx} = 144(1-32x)$
$\frac{dy}{dx} = \frac{-32x}{18y} = \infty$
Slope of normal = $-\frac{dx}{dy} = \frac{18y}{32x} = 0$
From this we can say that y = 0
Now. when y = 0,  $\frac{x^2 }{9} + \frac{0^2 }{16} \Rightarrow 1 = x = \pm 3$
Hence, the coordinates are (3,0) and (-3,0)

Exams
Articles
Questions