JEE Main 16 March 2021 Shift 1 Solved Paper

Given, equation of circle ⇒x2+y2=25
Equation of hyperbola ⇒‌

x2

9

−‌

y2

16

=1
Let the mid-point of the chord of the circle be (h,k).
The equation of a chord through its mid-point is simply T=S1
where T is the equation of tangent and S1 is the value of S by putting (h,k).
Here, ‌‌hx+ky=h2+k2
The equation of tangent to hyperbola ‌

x2

a2

−‌

y2

b2

=1 is
y=mx±√a2m2−b2
Here, y=mx±√9m2−16
So, y=‌

−hx

k

+‌

h2+k2

k

.....(i)
and y=mx±√9m2−16.....(ii)
Eqs. (i) and (ii) are identical, so
‌

1

1

=‌

−h∕k

m

=‌

h2+k2

k√9m2−16

⇒m=‌

−h

k

⇒9m2−16=‌

(h2+k2)2

k2

⇒9(‌

h2

k2

)−16=‌

(h2+k2)2

k2

⇒9h2−16k2=(h2+k2)2
{

h→x

k→y

So,