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

Step 1: Place the points.
Let $P(x,y)$ lie on $\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1$. The major axis is the $x$-axis with end vertices $A(-a,0)$ and $A'(a,0)$. Drop a perpendicular from $P$ to the axis, meeting it at $N(x,0)$.

Step 2: Read off the three lengths.
$PN$ is the vertical height, so $PN=y$. Also $AN=x-(-a)=x+a$ and $A'N=a-x$.

Step 3: Form the product $AN\cdot A'N$.
$AN\cdot A'N=(x+a)(a-x)=a^2-x^2$.

Step 4: Get $y^2$ from the ellipse.
From the ellipse equation, $\dfrac{y^2}{b^2}=1-\dfrac{x^2}{a^2}=\dfrac{a^2-x^2}{a^2}$, so $y^2=\dfrac{b^2}{a^2}(a^2-x^2)$.

Step 5: Form the required ratio.
$\dfrac{PN^2}{A'N\cdot AN}=\dfrac{y^2}{a^2-x^2}=\dfrac{\frac{b^2}{a^2}(a^2-x^2)}{a^2-x^2}$.

Step 6: Simplify.
The $(a^2-x^2)$ cancels, leaving $\dfrac{PN^2}{A'N\cdot AN}=\dfrac{b^2}{a^2}$. \[ \boxed{\dfrac{PN^2}{A'N\cdot AN}=\dfrac{b^2}{a^2}} \]