# Tangents are drawn from the point T(2,-1) to the parabola x^2=4y. P and Q are the points of contact of the tangents.a. Show the x-coordinates of P and Q are the roots of the quadratic x^2-4x-4=0

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`x^2 = 4y`

The gradient of a tangent line drawn to a curve is given by the first derivative of the curve.

Therefore gradient of the tangent line is dy/dx.

`x^2 = 4y`

Let us differentiate the above with respect to x.

2x = 4dy/dx

`(dy)/dx` = x/2

let us say the equation of the tangent line which passes through T(2,-1) is y = mx+c

Let us say the point of the curve which tangent goes through is (a,b).

Then m = dy/dx = x/2 = a/2 since tangent is at point P (a,b) on the curve.

So we can say;

-1 = m(2)+c

-1 = `a/2*2+c`

-1 = a+c

c = -1-a

Point (a,b) is on the curve. Therefore;

`a^2` = 4b

b = `a^2/4`

Now we substitute the point P (a,b) to the tangent line y = mx+c

`a^2/4` = `(a/2)*a-1-a`

`a^4/4` = `(a^2-2-2a)/2`

`a^2` = `2a^2-4-4a`

0 = `a^2-4a-4`

a represent the points on the tangent. Since it is quadratic equation both P and Q points will be given when solving it. If we replace **a** by **x**;

`0 = x^2-4x-4`

**So the coordinates of P and Q are the roots of quadratic `x^2-4x-4=0` **

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