Question

A normal is drawn to the parabola "y^(2)=9x" at the point "P(4,6)." A circle is described on "SP" as diameter; where "S" is the focus.The length of the intercept made by the circle on the normal at point "P" is:

Answer 1

Answer:

Please see the attachment

Answer 2

Answer is $\frac {15}{4}$

Step-by-step explanation:

Given, $y^2 = 9x$  

Comparing with standard equation of parabola : $y^2 = 4ax$

4a = 9 or a = $\frac {9}{4}$

hence, Value for $a = \frac {9}{4}$

Therefore, Focus = s($\frac {9}{4}$, 0)

As we know the coordinates of end points of diameter, the equation of circle can be given by:

$(x-x_1)(x-x_2) + (y-y_1)(y-y_2) = 0$

⇒ $(x-4)(x-\frac {9}{4}) + (y-6)(y-0) = 0$

⇒ $(x-4)(4x-9) + 4y(y-6) = 0$

⇒ $4x^2 + 4y^2-25x-24y+36 = 0$

⇒ $x^2 + y^2 - \frac {25}{4}x - 6y + 9 = 0$

Comparing with general equation of circle:  

$x^2 + y^2 +2gx + 2fy + c = 0$

Which on solving yields-

g = $\frac {-25}{8}$, f = -3, c = 9

Radius, r = $\sqrt {(\frac {25}{8})^2+9-9} = \frac {25}{8}$

Now right form of equation of normal :  

$y - y_1 = \frac {-y_1}{2a}(x-x_1)$

⇒ $(y-6) = \frac {-6}{\frac {9}{2}}(x-4)$

⇒ 3y-18 = -4x + 26

⇒ 4x + 3y -34 = 0 """ Equation of Normal

Therefore, Perpendicular distance of centre O ($\frac {-25}{8}$, 3) from normal = d

d = $\left |\frac {4 \times \frac {25}{8}+3(3)-34}{\sqrt {4^2+32}}\right |$

= $\left |\frac {\frac {25}{2}-25}{5}\right |$

d = $\frac {5}{2}$

We know perpendicular from center to chord bisects it

So, AB = AC

In ΔABO, By Pythagoras theorem,  

$AO^2 = OB^2 + AB^2$

$(\frac {25}{8})^2 = (\frac {5}{2})^2 + AB^2 = \frac {15}{4}$