Q.59. Tangents to Parabola from an external point

Question 59.
Find equations of both lines through the point (2,-3) that are tangent to the parabola y = x2 + x.

Answer 59.
Let the tangent line have slope m.
Then its equation is
y + 3 = m (x - 2)

If this line is a tangent to the given parabola, then it should intersect the parabola in two identical points which becomes the contact point of the tangent. Solving the two equations by eliminating y,
x^2 + x = m(x -2) - 3
x^2 - (m - 1)x + 2m + 3 = 0

For two values of x to be identical, discriminant should be zero.
=> (m - 1)^2 - 4(2m + 3) = 0
=> m^2 - 10m -11 = 0
=> (m - 11)(m + 1) = 0
=> m = 11 or - 1

Hence, equations of tangents are
y + 3 = 11(x - 2) and y + 3 = - 1(x - 2)
i.e., 11x - y = 25 and x + y + 1 = 0

Verification:
Putting m = 11 in the equation x^2 - (m - 1)x + 2m + 3 = 0
x^2 - 10x + 25 = 0
=> x = 5 and from equation of parabola, y = 30
Equation of tangent to the parabola at (x', y') is
(1/2) (y + y') = x'x + (1/2)(x + x')
Putting x' = 5 and y' = 30,
y + 30 = 10x + x + 5
=> 11x - y = 25
[This confirms that the equation of tangent obtained above is correct.]
Similarly, the equation of the second tangent can be verified to be correct.

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