Question

A parabola which touches x-axis at (1,0) and y-axis at (0,2). Then

• A. Vertex is $(\frac{16}{25},\frac{2}{25})$
• B. Vertex is $(\frac{16}{5},\frac{2}{5})$
• C. Focus is $(\frac{4}{5},\frac{2}{5})$
• D. Focus is $(1,\frac{1}{2})$

Answer 1

The correct option is D Focus is $(1,\frac{1}{2})$

We have,

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

Squaring both side and we get,

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

If passes through the point (1,0)

Then,

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

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

If passes through the point (0,2)

Then,

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

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

On adding equation (1) and (2) to and we get,

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

$h^2+1^2+k^2-2k+4+h^2-4h+k^2=\frac{1+m^2}{1+m^2}$

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

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

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

Hence the locus of (x, y)

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

Then the focus is $(-g,-f)=(1,\frac{1}{2})$

Hence, this is the answer.