Thank you for reporting, we will resolve it shortly
Q.
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
Application of Derivatives
Solution:
The equation of the given curve is $\frac{x^2}{9}+\frac{y^2}{16}=1 ....$(i)
On differentiating both sides w.r.t. $x$, we get
$\frac{2 x}{9}+\frac{1}{16}\left(2 y \frac{d y}{d x}\right)=0$
$\Rightarrow \frac{y}{8} \frac{d y}{d x}=-\frac{2 x}{9} \Rightarrow \frac{d y}{d x}=-\frac{16 x}{9 y} ....$(ii)
A. For tangent parallel to $X$-axis, we must have, $\frac{d y}{d x}=0$
$\Rightarrow -\frac{16 x}{9 y}=0 \Rightarrow x=0$
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{d x}{d y} =0 $
$\Rightarrow -\frac{9 y}{16 x} =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.