# Tangents are drawn from an external point P(2,-3) to the parabola x^2=4ay find the coordinate of the points at which each tangen meets the parabola

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### 1 Answer

The equation of the parabola is `x^2=4ay` or `y=x^2/(4a)` . We are drawing tangent lines through the point (2,-3) to the parabola.

The slopes of the tangent lines are given by the first derivative evaluated at the point of tangency.

`y=x^2/(4a)==>y'=x/(2a)`

Thus the equation of a tangent line to the parabola at `(x,x^2/(4a))` and through point (2,-3) is `y+3=x/(2a)(x-2)` or `y=x^2/(2a)-x/a-3` . But `y=(x^2)/(4a)` so

`x^2/(4a)=x^2/(2a)-x/a-3` Multiplying through by 4a we get:

`x^2=2x^2-4x-12a` or `x^2-4x-12a=0` . Using the quadratic formula we get:

`x=(4+-sqrt(16-4(1)(-12a)))/(2)`

`x=(4+-sqrt(16(1+3a)))/2`

`x=2+-2sqrt(1+3a)`

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The points where the tangents meet the parabola are:

`(2+2sqrt(1+3a),(2+2sqrt(1+3a)+3a)/(a))` and

`(2-2sqrt(1+3a),(2-2sqrt(1+3a)+3a)/a)`

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Examples: Let a=1 then the points of tangency are (6,9) and (-2,1). The tangent lines are y=3x-9 and y=-x-1

The graph:

If a=5 then the points of tangency are (10,5) and (`-6,9/5` ):