Finding tangents -calculus help needed please kinda urgent?

Question

i need help i cannot do this question so can somone help me

y = 9.35x2 Determine the positions on the curve (x,y), of all tangents that pass through the point (4.58,98.06). Enter none if no tangent exists.

ENTER THE LEFT-MOST TANGENT POINT (the one with the lowest x-value) IN THE FIRST BOX!

Round all values to FOUR decimal places, using the normal convention.

Left-most tangent point (x,y) =

Right-most tangent point (x,y) =

Answer 1

y = 9.35x^2

dy/dx = 18.7x

Therefore equation of tangent at the point (p, 9.35p^2) is

y - 9.35p^2 = 18.7p(x - p)

If this passes through (4.58, 98.06) then

98.06 - 9.35p^2 = 18.7p(4.58 - p)

9.35p^2 - 85.646p + 98.06 = 0

Try to solve this quadratic equation for two values of p. If there are any, you can find the coordinates from the p values.

Alternative method.

Any line through (4.58, 98.06) will have equation

y - 98.06 = m(x - 4.58)

This cuts the curve when

9.35x^2 - 98.06 = m(x - 4.58)

9.35x^2 - mx + (4.58m - 98.06) = 0

Trying to solve this equation will give the two x values of the points of intersection in terms of m. However, you don't need to do this. You want there to be only one solution for x so that the line is a tangent. This means that the discriminant (b^2 - 4ac) of the quadratic formula must be zero. Find what values of m (if any) make that true.