If the straight line lx+my=n is a normal to the hyperbola x2a2-y2b2=1, show that a2l2-b2m2=a2+b22n2 - Maths - Inverse Trigonometric Functions

Question

I f   t h e   s t r a i g h t   l i n e   l x +
m y = n   i s   a   n o r m a l   t o   t
h e   h y p e r b o l a   x 2 a 2 - y 2 b 2 = 1 ,  
s h o w   t h a t   a 2 l 2 - b 2 m 2 = a 2 + b 2 2 n 2

Neha Sethi · Accepted Answer

Dear student
Normals of slope m  to the hyperbola are given byy=Mx+Ma2+b2a2-b2M2So, for y=Mx+cc=Ma2+b2a2-b2M2c2=M2a2+b22a2-b2M2For lx+my-n=0⇒y=n-lxm⇒y=nm-lmx⇒y=-lmx+nmSo,c=nmand slope=-lmConsider,c2=M2a2+b22a2-b2M2⇒n2m2=l2m2a2+b22a2-b2l2m2⇒n2m2=l2a2+b22a2m2-b2l2⇒n2a2m2-b2l2=m2l2a2+b22⇒a2+b22n2=a2m2-b2l2m2l2⇒a2+b22n2=a2l2-b2m2
Regards

I f t h e s t r a i g h t l i n e l x + m y = n i s a n o r m a l t o t h e h y p e r b o l a x 2 a 2 - y 2 b 2 = 1 , s h o w t h a t a 2 l 2 - b 2 m 2 = a 2 + b 2 2 n 2 .

$I f t h e s t r a i g h t l i n e l x + m y = n i s a n o r m a l t o t h e h y p e r b o l a \frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1, s h o w t h a t \frac{a^{2}}{l^{2}} - \frac{b^{2}}{m^{2}} = \frac{{(a^{2} + b^{2})}^{2}}{n^{2}} .$