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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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