Question

If the equation of the curve is given by $\frac{x^2}{9}+\frac{y^2}{16}=1$, then the point, at which the tangents are

• A. parallel to $X$-axis, are $(0,\pm 4)$
• B. parallel to $Y$-axis, are $(\pm 4,0)$
• C. Both (a) and (b) are true
• D. Both (a) and (b) are false

Answer 1

Answer: (A)

The equation of the given curve is $\frac{x^2}{9}+\frac{y^2}{16}=\mathrm{1....}$(i)
On differentiating both sides w.r.t. $x$, we get
$\frac{2x}{9}+\frac{1}{16}\left(2y\frac{dy}{dx}\right)=0$
$\Rightarrow \frac{y}{8}\frac{dy}{dx}=-\frac{2x}{9}\Rightarrow \frac{dy}{dx}=-\frac{16x}{9y}....$(ii)
A. For tangent parallel to $X$-axis, we must have, $\frac{dy}{dx}=0$
$\Rightarrow -\frac{16x}{9y}=0\Rightarrow x=0$
When $x=0$, then from Eq (i), we get
$\frac{0^2}{9}+\frac{y^2}{16}=1\Rightarrow y^2=16\Rightarrow y=\pm 4$
Hence, the points on Eq. (i) at which the tangents are parallel to $X$-axis are $(0,4)$ and $(0,-4)$.
B. For tangents parallel to $Y$-axis, we must have,
$\frac{dx}{dy}=0$
$\Rightarrow -\frac{9y}{16x}=0$
$\Rightarrow y=0$
When, $y=0$, then from Eq (i), we get
$\frac{x^2}{9}+\frac{0^2}{16}=1\Rightarrow x^2=9\Rightarrow x=\pm 3$
Hence, the points on Eq. (i) at which the tangents are parallel to $Y$-axis are $(3,0)$ and $(-3,0)$.
So, only option (a) is correct.