Question

If tangents are drawn to the parabola $y^2=4ax$ at points whose abscissae are in the ratio $m^2:1$, then the locus of their point of intersection is the curve $(m>0)$

• A. $y^2=(m^{1/2}-m^{-1/2}{)}^{2}ax$
• B. $y^2=(m^{1/2}+m^{-1/2}{)}^{2}ax$
• C. $y^2=(m^{1/2}+m^{-1/2}{)}^{2}x$
• D. $y^2=(m^{1/2}-m^{-1/2}{)}^{2}x$

Answer 1

Answer: (B)

$y^2=(m^{1/2}+m^{-1/2}{)}^{2}ax$

Let one of the points be $(a{t}^2,2at)$, equation of tangent at this point is

$2aty=2a(a{t}^2+x)$

$ty=x+a{t}^2--\left(1\right)$

(slope is $\frac{1}{t}$ and abscissa $=a{t}^2$)

Let abscissa of other point be $t_1$

Ratio of abscissa is $\frac{m^2}{1}$

So, $\frac{a{t}^2}{a{t}_1^2}=m^2$

So, $t_1=\frac{t}{m}$

Equation of tangent at the other point
$\frac{2aty}{m}=2a(x+\frac{a{t}^2}{m^2})$
$\frac{ty}{m}=x+\frac{a{t}^2}{m^2}--\left(2\right)$

To find the point of intersection, subsitute $x$ from equation $\left(1\right)$ in equation $\left(2\right)$
$yt(1-\frac{1}{m})=a{t}^2(1-\frac{1}{m^2})$
$y=at(1+\frac{1}{m})--\left(3\right)$
$a+2+\frac{a+2}{m}=x+a{t}^2$
$x=\frac{a{t}^2}{m}--\left(4\right)$
By elliminating $t$ in $\left(3\right)$ and $\left(4\right)$
$y^2=(m^{\frac{1}{2}}+m^{\frac{-1}{2}}{)}^{2}ax$