Question

Two tangents are drawn to end points of the latus rectum of the parabola $y^2=4x$. The equation of the parabola which touches both the tangents as well as the latus rectum is

• A. $y^2=8x$
• B. $y^2=8(x-1)$
• C. $y^2=4(x-1)$
• D. $(y-1{)}^2=8x$

Answer 1

The correct option is B $y^2=8(x-1)$

The end points of the latus rectum are $(1,\pm 2)$
equation of the tangents drown to the end points are
$y=\pm (x+1)\cdots \left(1\right)$
By the symmetry, it can be concluded as axis of the required parabola will be same as the axis of the given parabola.
Now let the equation of the required parabola is
$y^2=4a(x-1)\cdots \left(2\right)$
equating $\left(1\right)$ and $\left(2\right)$
$x^2+1+2x=4a(x-1)$
For tangency $D=0$
$\Rightarrow (2-4a{)}^2=4(4a+1)\Rightarrow 4+16a^2-16a=16a+4\Rightarrow a=0,2$
$a=0$ is not valid
So equation will be $y^2=8(x-1)$