Question

If a normal is drawn at a variable point P(x, y) on the curve $9x^2+16y^2-144=0$, then the maximum distance from the centre of the curve to the normal is

• A. 1
• B. 7
• C. 12
• D. $\frac{3}{4}$

Hint:

For an ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ (with $a>b$), the maximum distance from the center to a normal is $a-b$, and the minimum distance is $a-b$. For a hyperbola, the minimum distance from the center to a normal is $a+b$. Remembering these standard results can provide an instant answer to such questions.

Answer 1

The Correct Option is A

Step 1: Identify the curve \( 9x^2 + 16y^2 = 144 \implies \frac{x^2}{16} + \frac{y^2}{9} = 1 \). This is an ellipse with \( a^2 = 16 \implies a=4 \) and \( b^2 = 9 \implies b=3 \).
Step 2: Equation of the Normal The equation of the normal at a point \( (a\cos\theta, b\sin\theta) \) is given by: \( ax \sec\theta - by \csc\theta = a^2 - b^2 \). Substitute \( a=4, b=3 \): \( 4x \sec\theta - 3y \csc\theta = 16 - 9 = 7 \).
Step 3: Distance from Center The center of the ellipse is \( (0,0) \). The perpendicular distance \( d \) from the origin to the normal is: \[ d = \frac{| -7 |}{\sqrt{(4\sec\theta)^2 + (-3\csc\theta)^2}} = \frac{7}{\sqrt{16\sec^2\theta + 9\csc^2\theta}} \]
Step 4: Maximize the Distance To maximize \( d \), we must minimize the denominator \( D = 16\sec^2\theta + 9\csc^2\theta \). Using the identity \( \sec^2\theta = 1+\tan^2\theta \) and \( \csc^2\theta = 1+\cot^2\theta \): \( D = 16(1+\tan^2\theta) + 9(1+\cot^2\theta) = 25 + 16\tan^2\theta + 9\cot^2\theta \). Using AM-GM inequality for the variable part: \( 16\tan^2\theta + 9\cot^2\theta \ge 2\sqrt{16 \cdot 9} = 2(12) = 24 \). So \( D_{min} = 25 + 24 = 49 \). Alternatively, the minimum of \( A\sec^2\theta + B\csc^2\theta \) is \( (\sqrt{A} + \sqrt{B})^2 \). \( D_{min} = (\sqrt{16} + \sqrt{9})^2 = (4+3)^2 = 49 \).
Step 5: Final Calculation Maximum Distance \( = \frac{7}{\sqrt{49}} = \frac{7}{7} = 1 \).