Differentiating curve equation
25x4 – 30x2 + 1 + 2m =0
2m= 30x2 - 25x4 – 1
m =15x2 – (25/2)x4 – 1/2
m(0,-3) = -1/2
So slope of normal at p(0, -3) = -1/m(0,-3) =2
It is also tangent to curve at some points (x,y) then
m(x,y) = 2 =15x2 – (25/2)x4 – ½
So 3 = 30x2 – 25x4
25x4 - 30x2 +3 = 0
Solve this equation for value of x and then put value of x in equation of curve to find value of y.
						

							
Dear Aayushi
 
Let us assume the question as: The normal to the curve 5x5 - 10x3 + x + 2y + 6 = 0 at point (0, -3) is tangent to the curve at some other point. Find the point ?

To solve this, we first have to find the equation of normal to the given the given curve at the point (0, -3). For that, we need to differentiate the equation of the curve and find the gradient. The gradient of the given curve will be: 
 
 
25 x⁴ - 30 x² +1 + 2dy/dx = 0
 
Now value of gradient at (0 , -3)
dy/dx = -1/2
 
Let as assume the value as m1 = -1/2

We also know that slope of the normal will be: m2 = (-1 / m1). Hence, m2 = 2.

So, equation of the normal can be found by using point slope form:

(y - y1) = m2 (x - x1)
(y - (-3)) = 2 (x - 0)
y + 3 = 2x
y = 2x - 3 ---------- (i)

According to the question, equation (i) is also tangent to the given curve. We can plug in the value of y from equation (i) into the equation of the curve. So, we get: 

5x​5 - 10x3 + x + 2(2x - 3) + 6 = 0
5x5 - 10x3 + x + 4x - 6 + 6 = 0 
5x5 - 10x3 + 5x = 0
5x(x4 - 2x2 + 1) = 0
x4 - 2x2 + 1 = 0
x4 - x2 - x2 + 1 = 0
x2 (x2 - 1) - 1(x2 - 1) = 0
(x2 - 1)2 = 0
x = 1

Plugging in the value of x = 1 in equation (i). We get, y = 2 - 3 = -1. 

Hence, the required point is (1, -1). 
 
 
Regards
Arun (askIITians forum expert)