Answer to Question #221384 in Calculus for Astrics

Question

the normal to a parabola y^2=4ax at the point t,where t is not 0 meets the curve again at the point t'.find t' in terms of t.determine a point on the x-axis where the tangent at t' meets the x-axis.

Answer 1

Parabola : $y^2 = 4ax$

Any point P on the parabola in parametric form is given as : $(at^2, 2at)$

The slope of tangent at any point is $\dfrac{1}{t}$

Hence, the slope of normal at point $(at^2, 2at)$ is = $-t$

The equation of the normal becomes

$y - 2at = -t \ (x - at^2)$

$y = -t x + at^3 + 2at$

If the above equation intersects the curve again at point at ($at'^2, 2at'$ )

Then it would satisfy the equation of normal. On substituting the value we get

$2at' = -att'^2 + at^3 + 2at$

$2a(t' - t) = at(t-t')(t+t')$

$2a = -at(t+t')$

$\dfrac{-2}{t} = t + t'$

$t' = \dfrac{-2}{t} - t$

Hence, the normal to the parabola meets the curve again at point $t' = \dfrac{-2}{t} - t$

Now, let us find the tangent to the curve at the point (at'² , 2at' )

Slope of tangent at this point is $\dfrac{1}{t'}$

Equation of tangent is given by

$y - 2at' = \dfrac{1}{t'}(x - at'^2)$

To find the point where the above tangent meets the x-axis, we put y = 0.

Hence on substituting y = 0

$- 2at' = \dfrac{1}{t'}(x - at'^2)$

$x= -at'^2$

Hence, the tangent intersects the curve at ( $-at'^2,0$ ).