Question

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

• A. (0,4), (0,-4)
• B. (0, 0), (0,-4)
• C. (4, 0), (-4, 0)
• D. (0, 0), (0, 4)

Answer 1

The Correct Option is A

To find the points where the tangents to the curve \(\frac{x^2}{9}+\frac{y^2}{16}=1\) are parallel to the x-axis, we need to find where the derivative \(\frac{dy}{dx}=0\).

Given the curve equation:

\[\frac{x^2}{9}+\frac{y^2}{16}=1\]

Differentiating implicitly with respect to \(x\):

\[\frac{d}{dx}\left(\frac{x^2}{9}\right)+\frac{d}{dx}\left(\frac{y^2}{16}\right)=0\]

Apply chain rule:

\[\frac{2x}{9}+\frac{2y}{16}\cdot\frac{dy}{dx}=0\]

Simplify and solve for \(\frac{dy}{dx}\):

\[\frac{2y}{16}\cdot\frac{dy}{dx}=-\frac{2x}{9}\]

\[\frac{dy}{dx}=-\frac{2x}{9}\cdot\frac{16}{2y}\]

\[\frac{dy}{dx}=-\frac{16x}{9y}\]

For the tangent to be parallel to the x-axis, \(\frac{dy}{dx}=0\). This occurs when the numerator is zero:

\[-16x=0\]

Solving gives:

\[x=0\]

Substituting \(x=0\) back into the original curve equation:

\[\frac{0^2}{9}+\frac{y^2}{16}=1\]

\[0+\frac{y^2}{16}=1\]

Multiply by 16:

\[y^2=16\]

Taking the square root:

\[y=\pm 4\]

So, the points where the tangents are parallel to the x-axis are:

(0,4) and (0,-4).

Thus, the correct option is:

(0,4), (0,-4)