Question

If \(P\) is any point on the ellipse \[ \frac{x^2}{a^2}+\frac{y^2}{b^2}=1 \] with major axis \(AA'\), and \(N\) is the foot of the perpendicular drawn from \(P\) upon \(AA'\), then

• A. \[ \frac{PN^2}{A'N+AN} = \frac{b^2}{a^2} \]
• B. \[ \frac{PN^2}{A'N+AN} = \frac{a^2}{b^2} \]
• C. \[ \frac{PN^2}{A'N\cdot AN} = \frac{b^2}{a^2} \]
• D. \[ \frac{PN^2}{A'N\cdot AN} = \frac{a^2}{b^2} \]

Hint:

For the ellipse \[ \frac{x^2}{a^2}+\frac{y^2}{b^2}=1, \] a useful result is \[ PN^2 = \frac{b^2}{a^2} (AN)(A'N). \] This is frequently used in coordinate geometry problems.

Answer 1

The Correct Option is C

Concept: Let \[ P(x,y) \] be any point on the ellipse \[ \frac{x^2}{a^2}+\frac{y^2}{b^2}=1. \] The major axis is the \(x\)-axis with vertices \[ A(-a,0),\qquad A'(a,0). \]

Step 1: Find \(PN\), \(AN\) and \(A'N\). Foot of perpendicular: \[ N(x,0) \] Therefore, \[ PN=y \] \[ AN=x+a \] \[ A'N=a-x \]

Step 2: Find the product \(AN\cdot A'N\). \[\begin{aligned} AN\cdot A'N &= (x+a)(a-x) \\ &= a^2-x^2 \end{aligned}\]

Step 3: Use the ellipse equation. \[ \frac{x^2}{a^2}+\frac{y^2}{b^2}=1 \] \[ y^2 = b^2\left(1-\frac{x^2}{a^2}\right) \] \[ y^2 = \frac{b^2}{a^2}(a^2-x^2) \] Since \[ PN=y, \] \[ PN^2 = \frac{b^2}{a^2}(AN\cdot A'N) \] Hence, \[ \frac{PN^2}{AN\cdot A'N} = \frac{b^2}{a^2} \] \[\begin{aligned} \boxed{ \frac{PN^2}{AN\cdot A'N} = \frac{b^2}{a^2} } \end{aligned}\] Hence, option \(\mathbf{(C)}\) is correct.