Question

Parabola which touches x axis at (1,0) and y axis as (0,2)

Answer 1

Answer:

          Given in the question,  Parabola touches X-axis at (1,0)  and Y-axis at (0,2).

            Let the equation of directrix is parabola y=mx  and co-ordinate focus of parabola is (h,k).

We know that, Equation of parabola is

$\sqrt{(x-h)^{2} +(y-k)^{2} }$ = / $\frac{y-mx}{\sqrt{1+m^2} }} }$ /

Taking squares on both Sides, we get

${(x-h)^{2} +(y-k)^{2} }$ = /  $\frac{(y-mx)^2}{(1+m)^{2} }$  /

If passes through the point (1,0)  Then,

$(0-h)^{2}$ + $(1-k)^{2}$ = $\frac{1}{1+m^2}$   .............. equation 1

It also passes through (0,2)

$(2-h)^{2}$ + $k^2$ = $\frac{m^2}{1+m^2}$   .................  equation 2

Adding equation 1 and 2, we have

$h^2$ + $(1-k)^{2}$ + $(2-h)^{2}$ + $k^2$ =  $\frac{1}{1+m^2}$  +  $\frac{m^2}{1+m^2}$

2$h^2$ + 2$k^2$ - 4h - 2k + 5 = 1

2$h^2$ + 2$k^2$ - 4h - 2k + 4 = 0

$h^2$ + $k^2$ - 2h - k + 2 = 0

Hence the locus of (x, y)

$x^2 + y^2 - 2x - y + 2 = 0$

Hence, The Locus is (-g,-f) = ( 1 , $\frac{1}{2}$ )

Answer 2

Given :

The parabola touches X-axis at point A(1,0)

The parabola touches Y-axis at point B(0,2)

To Find :

The equation of parabola

Solution :

• The general equation of parabola having focus(h,k) and directrix equation y=mx is

                $\sqrt{(x-h)^{2}+(y-k)^{2} } = \frac{y-mx}{\sqrt{1+m^{2} } }$

     

                 $(x-h)^{2}+(y-k)^{2} = \frac{y-mx ^{2} }{1+m^{2} } }$

• The parabola passes through the point A(0,1)

                $(0-h)^{2}+(1-k)^{2} = \frac{1 }{1+m^{2} } } \rightarrow Equation (1)$

• The parabola passes through the point B(2,0)

                $(2-h)^{2}+(0-k)^{2} = \frac{m^{2} }{1+m^{2} } }\rightarrow Equation(2)$

• By adding equations (1) and (2) we get

            $(0-h)^{2}+(1-k)^{2} + (2-h)^{2}+(0-k)^{2} = \frac{1 }{1+m^{2} } }+ \frac{m^{2} }{1+m^{2} } }$

            $2h^{2} + 2k^{2} -4h -2k +4 =0$

            $h^{2} +k^{2} -2h-k+2=0$      

The locus of the equation of parabola is  x²+y²-2x-y+2=0