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: Understanding the Concept
The problem asks for the maximum possible perpendicular distance from the center of an ellipse to any of its normal lines. We need to find the equation of a general normal, calculate its distance from the center, and then maximize this distance.
Step 2: Key Formula or Approach
1. Standardize the equation of the ellipse to find its semi-major ($a$) and semi-minor ($b$) axes. The center of this ellipse is the origin (0,0). 2. The equation of the normal to the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ at a point with eccentric angle $\theta$ is $ax\sec\theta - by\csc\theta = a^2-b^2$. 3. The perpendicular distance from the origin $(0,0)$ to a line $Ax+By+C=0$ is given by $d = \frac{|C|}{\sqrt{A^2+B^2}}$. 4. We will express this distance $d$ as a function of $\theta$ and find its maximum value. This is a standard result for an ellipse.
Step 3: Detailed Explanation
1. Ellipse parameters: The equation of the curve is $9x^2+16y^2=144$. Dividing by 144, we get the standard form: \[ \frac{9x^2}{144} + \frac{16y^2}{144} = 1 \implies \frac{x^2}{16} + \frac{y^2}{9} = 1 \] Comparing with $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$, we have $a^2=16 \implies a=4$ and $b^2=9 \implies b=3$. The center of the ellipse is $(0,0)$. 2. Equation of the normal: The equation of the normal at the point $(4\cos\theta, 3\sin\theta)$ is: \[ 4x\sec\theta - 3y\csc\theta = 4^2 - 3^2 = 16 - 9 = 7 \] In the form $Ax+By+C=0$, this is $\left(\frac{4}{\cos\theta}\right)x - \left(\frac{3}{\sin\theta}\right)y - 7 = 0$. 3. Distance from the center to the normal: The distance $d$ from the center $(0,0)$ to this normal line is: \[ d = \frac{|-7|}{\sqrt{\left(\frac{4}{\cos\theta}\right)^2 + \left(-\frac{3}{\sin\theta}\right)^2}} = \frac{7}{\sqrt{16\sec^2\theta + 9\csc^2\theta}} \] 4. Maximize the distance: To maximize $d$, we must minimize the denominator, $D = \sqrt{16\sec^2\theta + 9\csc^2\theta}$. This is equivalent to minimizing $D^2 = 16\sec^2\theta + 9\csc^2\theta$. Using trigonometric identities $\sec^2\theta = 1+\tan^2\theta$ and $\csc^2\theta = 1+\cot^2\theta$: \[ D^2 = 16(1+\tan^2\theta) + 9(1+\cot^2\theta) = 16+16\tan^2\theta + 9+9\cot^2\theta \] \[ D^2 = 25 + 16\tan^2\theta + 9\cot^2\theta \] Let $f(\theta) = 16\tan^2\theta + 9\cot^2\theta$. We can find the minimum value using AM-GM inequality or calculus. Using AM-GM: \[ \frac{16\tan^2\theta + 9\cot^2\theta}{2} \ge \sqrt{(16\tan^2\theta)(9\cot^2\theta)} = \sqrt{144} = 12 \] So, $16\tan^2\theta + 9\cot^2\theta \ge 24$. The minimum value of $D^2$ is $25 + 24 = 49$. The minimum value of the denominator $D$ is $\sqrt{49} = 7$. Therefore, the maximum value of the distance $d$ is: \[ d_{max} = \frac{7}{D_{min}} = \frac{7}{7} = 1 \] A standard result for ellipses: The maximum distance from the center to any normal is $a-b$. Here, $a=4$ and $b=3$, so the maximum distance is $4-3=1$. Step 4: Final Answer
The maximum distance from the centre of the ellipse to the normal is 1.