JEE Main Math Class 12 Application of Derivatives Part 1 Questions

Given, 4x2−9y2=36
After differentiating w.r.t. x, we get
4.2.x−9.2.y.

dy

dx

=0
⇒ Slope of tangent =

dy

dx

=

4x

9y

So, slope of normal =

−9y

4x

Now, equation of normal at point (x0,y0) is given byy−y0=

−9y0

4x0

(x−x0)
As normal intersects X axis at A, Then
A≡(

13x0

9

,0) and B≡(0,

13y0

4

)
As OABP is a parallelogram
midpoint of OB≡(0,

13y0

8

)≡ Midpoint of AP
So, P(x,y)≡(

−13x0

9

,

13y0

4

)
∵(x0,y0) lies on hyperbola, therefore
4(x0)2−9(y0)2=36 .......(ii)
From equation (i): x0=

−9x

13

and y0=

4y

13

From equation (ii), we get
9x2−4y2=169
Hence, locus of point P is :9x2−4y2=169