Strange Perpendicular Chords

Since it is independent of orientation we can WLOG assume the parabola to be $y^2 = 4ax$ . Its vertex is the origin.

(Refer to the image above.)

Consider two mutually perpendicular chords of lengths $r_1$ and $r_2$ respectively. Hence points on the parabola can be written as $(r_1 \cos \theta , r_1 \sin \theta )$ and $(r_2 \sin \theta , r_2 \cos \theta )$ . Naturally they must satisfy the equation of the parabola, thereby yielding us the following equations:

$\displaystyle r_1 = \frac{4a \cos \theta }{ \sin ^2 \theta }$

$\displaystyle r_2 = \frac{4a \sin \theta }{ \cos ^2 \theta }$

Now we need to simply eliminate $\theta$ . Consider,

$\displaystyle r_1^{\frac{2}{3}} +r_2^{\frac{2}{3}} = (4a)^{\frac{2}{3}} \bigg( \frac{ \cos ^2 \theta + \sin ^2 \theta }{\sin ^2 \theta \cos ^2 \theta } \bigg)^{\frac{2}{3}}$

$\displaystyle \Rightarrow r_1^{\frac{2}{3}} +r_2^{\frac{2}{3}} = \frac{1}{16a^2} \bigg( \frac{16a^2}{\sin \theta \cos \theta } \bigg)^{\frac{4}{3}}$ .

$\displaystyle \Rightarrow \boxed{ (r_1 r_2)^{\frac{4}{3}} = |\text{LR}|^2 ( r_1^{\frac{2}{3}} + r_2^{\frac{2}{3}} ) }$

where, $|\text{LR}| = 4a =$ Length of Latus Rectum.

Substitute $r_1 =8$ and $r_2 = 27$ to obtain $|\text{LR}| = \frac{36}{\sqrt{13}} \Rightarrow a+b = \boxed{49}$

I solved it in this way : , Let standard equation of Parabola as : ${ y }^{ 2 }=4ax$ Vertex is Origin O(0,0) and Let two chord's OA and OB , Also let Co-ordinates of A & B in paramateric form as : $A({ t }_{ 1 })\quad ,\quad B({ t }_{ 2 })$ . and again let length of chord's are ${ { l }_{ 1 } }\quad ,\quad { l }_{ 2 }$ .

Using ${ m }_{ 1 }{ m }_{ 2 }=-1\quad \quad \Rightarrow \quad \boxed { { t }_{ 1 }{ t }_{ 2 }\quad =\quad -4\quad } \quad (1)$ . Now Using distance formula for OA and OB : ${ { (a{ t }_{ 1 } }^{ 2 }) }^{ 2 }\quad +\quad { (2a{ t }_{ 1 }) }^{ 2 }\quad =\quad { { (l }_{ 1 } })^{ 2 }\quad \quad .\quad .\quad .\quad (2)\\ \\ { { (a{ t }_{ 2 } }^{ 2 }) }^{ 2 }\quad +\quad { (2a{ t }_{ 2 }) }^{ 2 }\quad =\quad { { (l }_{ 2 } })^{ 2 }\quad \quad .\quad .\quad .\quad (3)$ .

we have 3 equation's and 3 variables , so on simplyfying it By making quadratic in ' $t^2$ ' we get :

$\boxed { { (-{ t }_{ 1 }{ t }_{ 2 }) }^{ 2 }=\quad (\cfrac { \sqrt { { 4a }^{ 2 }+{ { { (l }_{ 1 } }) }^{ 2 } } -2a }{ a } )(\cfrac { \sqrt { { 4a }^{ 2 }+{ { { (l }_{ 2 } }) }^{ 2 } } -2a }{ a } ) } \quad (4)$ .

Now Solve this we should get

$a=\cfrac { 9 }{ \sqrt { 13 } } \quad \Rightarrow \quad \boxed { LR\quad =\quad \cfrac { 36 }{ \sqrt { 13 } } } \quad Ans.$ .

i too used polar coordinates..but in a easier way.from sudeep salgia 's answer we know r1 and r2 in terms of (theta).Take r1/r2 and u will find tan(theta)=2/3 (i hope i am right). Now u can find sin(theta) and cos(theta) in terms of tan and simply put it into one of the eqn's and get 4a=latus rectum