If the curves frac{x^2}{4}+frac{y^2}{9}=1 and | vedclass

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(D) Given curves are $\frac{x^2}{4}+\frac{y^2}{9}=1$ $(i)$ and $\frac{x^2}{16}-\frac{y^2}{k}=1$ (ii). For curve $(i)$, differentiating with respect to

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$-21$

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(D) Given curves are $\frac{x^2}{4}+\frac{y^2}{9}=1$ $(i)$ and $\frac{x^2}{16}-\frac{y^2}{k}=1$ (ii). For curve $(i)$, differentiating with respect to $x$: $\frac{2x}{4} + \frac{2yy'}{9} = 0 \Rightarrow y'_1 = -\frac{9x}{4y}$. For curve (ii), differentiating with respect to $x$: $\frac{2x}{16} - \frac{2yy'}{k} = 0 \Rightarrow y'_2 = \frac{kx}{16y}$. Since the curves are orthogonal, $y'_1 	imes y'_2 = -1$. Substituting the derivatives: $(-\frac{9x}{4y}) 	imes (\frac{kx}{16y}) = -1 \Rightarrow \frac{9kx^2}{64y^2} = 1 \Rightarrow 9kx^2 = 64y^2$. From $(i)$, $y^2 = 9(1 - \frac{x^2}{4}) = \frac{9(4-x^2)}{4}$. Substitute $y^2$ into the orthogonality condition: $9kx^2 = 64 	imes \frac{9(4-x^2)}{4} = 16 	imes 9(4-x^2) = 144(4-x^2)$. $kx^2 = 16(4-x^2) = 64 - 16x^2 \Rightarrow x^2(k+16) = 64 \Rightarrow x^2 = \frac{64}{k+16}$. Substitute $x^2$ into $(i)$: $\frac{64}{4(k+16)} + \frac{y^2}{9} = 1 \Rightarrow \frac{16}{k+16} + \frac{y^2}{9} = 1 \Rightarrow \frac{y^2}{9} = 1 - \frac{16}{k+16} = \frac{k}{k+16} \Rightarrow y^2 = \frac{9k}{k+16}$. Substitute $x^2$ and $y^2$ into $9kx^2 = 64y^2$: $9k(\frac{64}{k+16}) = 64(\frac{9k}{k+16})$. This is an identity for any $k$ where the intersection exists. Subtracting the two original equations: $(\frac{1}{4} - \frac{1}{16})x^2 + (\frac{1}{9} + \frac{1}{k})y^2 = 0 \Rightarrow \frac{3}{16}x^2 + \frac{k+9}{9k}y^2 = 0$. Using $9kx^2 = 64y^2 \Rightarrow x^2 = \frac{64y^2}{9k}$, substitute into the subtraction: $\frac{3}{16}(\frac{64y^2}{9k}) + \frac{k+9}{9k}y^2 = 0 \Rightarrow \frac{4y^2}{3k} + \frac{(k+9)y^2}{9k} = 0$. Dividing by $y^2/9k$: $12 + k + 9 = 0 \Rightarrow k = -21$.

Column-$I$	Column-$II$
$A$. Circle	$P$. Locus of point $(h, k)$ such that the line $hx + ky = 1$ touches the circle $x^2 + y^2 = 4$
$B$. Parabola	$Q$. Point $z$ in the complex plane satisfies $\|z + 2\| - \|z - 2\| = \pm 3$
$C$. Hyperbola	$R$. Eccentricity of the conic lies in the interval $[1, \infty)$
	$S$. Point $z$ in the complex plane satisfies $Re(z + 1)^2 = \|z\|^2 + 1$

If the curves $\frac{x^2}{4}+\frac{y^2}{9}=1$ and $\frac{x^2}{16}-\frac{y^2}{k}=1$ cut each other orthogonally, then $k=$

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