If a normal is drawn at a variable point P(x, | vedclass

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(A) The given equation of the curve is $9x^2 + 16y^2 = 144$. Dividing by $144$,we get $\frac{x^2}{16} + \frac{y^2}{9} = 1$. This is an ellipse with $a

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(A) The given equation of the curve is $9x^2 + 16y^2 = 144$. Dividing by $144$,we get $\frac{x^2}{16} + \frac{y^2}{9} = 1$. This is an ellipse with $a^2 = 16$ and $b^2 = 9$,so $a = 4$ and $b = 3$. The centre of the ellipse is $(0, 0)$. The equation of the normal at any point $(x_1, y_1)$ on the ellipse is $\frac{a^2x}{x_1} - \frac{b^2y}{y_1} = a^2 - b^2$. Substituting $a^2 = 16$ and $b^2 = 9$,we get $\frac{16x}{x_1} - \frac{9y}{y_1} = 7$. The distance $d$ from the origin $(0, 0)$ to the line $Ax + By + C = 0$ is given by $d = \frac{|C|}{\sqrt{A^2 + B^2}}$. Here,the line is $\frac{16}{x_1}x - \frac{9}{y_1}y - 7 = 0$. Thus,$d = \frac{|-7|}{\sqrt{(\frac{16}{x_1})^2 + (\frac{9}{y_1})^2}} = \frac{7}{\sqrt{\frac{256}{x_1^2} + \frac{81}{y_1^2}}}$. Since $x_1 = 4\cos	heta$ and $y_1 = 3\sin	heta$,we have $d = \frac{7}{\sqrt{\frac{256}{16\cos^2	heta} + \frac{81}{9\sin^2	heta}}} = \frac{7}{\sqrt{16\sec^2	heta + 9\csc^2	heta}}$. To maximize $d$,we must minimize $f(	heta) = 16\sec^2	heta + 9\csc^2	heta = 16(1 + 	an^2	heta) + 9(1 + \cot^2	heta) = 25 + 16	an^2	heta + 9\cot^2	heta$. By $AM$-$GM$ inequality,$16	an^2	heta + 9\cot^2	heta \ge 2\sqrt{16 	imes 9} = 2 	imes 12 = 24$. The minimum value of $f(	heta)$ is $25 + 24 = 49$. Thus,the maximum distance is $d_{max} = \frac{7}{\sqrt{49}} = \frac{7}{7} = 1$.

If a normal is drawn at a variable point $P(x, y)$ on the curve $9x^2 + 16y^2 = 144$,then the maximum distance from the centre of the curve to the normal is

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