Question

If the normal at any point $P$ on the ellipse $\frac{x^2}{25}+\frac{y^2}{16}=1$ with centre $C$ meets the major and minor axes at $M$ and $m$ respectively, and if $CN$ be perpendicular upon this normal, then

• A. the value of $PN\cdot PM$ is equal to $16.$
• B. the value of $PN\cdot Pm$ is equal to $25.$
• C. the value of $PM\cdot Pm$ is equal to $20$ if eccentric angle of the point $P$ is $\frac{\pi }{4}.$
• D. eccentricity of the ellipse is $\frac{3}{5}.$

Answer 1

Answer: (A), (B), (D)

The correct options are
A the value of $PN\cdot PM$ is equal to $16.$
B the value of $PN\cdot Pm$ is equal to $25.$
D eccentricity of the ellipse is $\frac{3}{5}.$

Let point $P(x_1,y_1)=(a\cos \theta ,b\sin \theta )$ on ellipse.
Equation of normal is $\frac{a^{2}x}{x_1}-\frac{b^{2}y}{y_1}=a^2{e}^2$

$\left(i\right)$ $PN\cdot PM=b^2$
LHS $=$ Power of the point $P$ with respect to the circle on $CM$ as diameter
$x_1(x_1-e^2{x}_1)+y_1^2$ $\{$ using diametric form of circle $\}$
$=x_1^2(1-e^2)+y_1^2$
$=a^2{\cos }^2\theta (1-1+\frac{b^2}{a^2})+b^2{\sin }^2\theta$

$=b^2{\cos }^2\theta +b^2{\sin }^2\theta$
$=b^2$

$\left(ii\right)$ $PN\cdot Pm=a^2$
LHS $=$ Power of the point $P$ with respect to the circle on $Cm$ as diameter
$=x_1^2+y_1(y_1+\frac{a^2{e}^2{y}_1}{b^2})$
$=a^2{\cos }^2\theta +b^2{\sin }^2\theta (1+\frac{(a^2-b^2)}{b^2})$
$=a^2{\cos }^2\theta +b^2{\sin }^2\theta \cdot \frac{a^2}{b^2}=a^2$

$\left(iii\right)$ $PM\cdot Pm=SP\cdot S^{\prime }P$
$RHS=(a-ae\cos \theta )(a+ae\cos \theta )$
$a^2-a^2{e}^2{\cos }^2\theta$
$a^2-(a^2-b^2){\cos }^2\theta$
$a^2{\sin }^2\theta +b^2{\cos }^2\theta$
LHS $=$ Power of $P$ with respect to
the circle on $Mm$ as diameter

$=x_1(x_1-e^2{x}_1)+y_1(y_1+\frac{a^2{e}^2{y}_1}{b^2})$
$=x_1^2(1-e^2)+y_1^2(1+\frac{a^2{e}^2}{b^2})$
$=x_1^2\left(\frac{b^2}{a^2}\right)+y_1^2(1+\frac{a^2-b^2}{b^2})$
$=b^2{\cos }^2\theta +a^2{\sin }^2\theta =RHS$