Find the equation of tangents to the parabola drawn from an external point P?

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if (x,y) is the point then y*y - 4ax is the tangent to the parabola.
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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...
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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). read less
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