Question

The locus of the point of intersection of tangents to $y^2=4ax$ which intercept a constant length $d$ on the directrix is

• A. $(y^{}-4ax){(x+a)}^2=d^2{x}^2$
• B. ${(y^{}-4ax)}^2{(x+a)}^2=d^2{x}^2$
• C. $y^2-4ax=d^2{x}^2{(x+a)}^2$
• D. None of these

Answer 1

Answer: (A)

$(y^{}-4ax){(x+a)}^2=d^2{x}^2$

Let $P$ be the point of intersection of the tangents to the parabola .

The equation of the pair of tangents. $PQ$ and $PR$ is

${\left[\left(y{y}_1\right)-2a\left(x+x_1\right)\right]}^2=\left(y^2-4^{a}x\right)\left(y_1^2-4a{x}_1\right)$

These lines meet the directix $x+1=0$ or $x=-a$.

Substituting this we get,

${\left[\left(y{y}_1\right)-2a\left(-a+x_1\right)\right]}^2=\left(y^2-4a^2\right)\left(y_1^2-4a{x}_1\right)$

On expanding we get,

${\left(y{y}_1\right)}^2+{\left[2a\left(x_1-a\right)\right]}^2-4ay{y}_1\left(x-a\right)={\left(y{y}_1\right)}^2-4a{x}_1{y}^2+4a{x}_1{y}^2+4a^2{y}_1^2-16a^3{x}_1$

(i.e) $y^2+\left[y_1^2-y_1^2+4a{x}_1\right]-y\left[4a\left(x_1-a\right)y_1\right]+\left[4a^2{\left(x_1-a\right)}^2-4a^2\left(y_1^2-4a{x}_1\right)y\right]=0$

$x_1{y}^2-y_1\left(x_1-a\right)y+a\left[{\left(x_1-a\right)}^2-\left(y_1^2-4a{x}_1\right)\right]=0$

$\Rightarrow x_1{y}^2-y{y}_1\left(x_1-a\right)+a\left[{\left(x_1-a\right)}^2-y_1^2\right]=0$

Here $y_1$ and $y_2$ are the ordiantes of the point of intersection of tangents with directix $x+a=0$.

Sum of the ordinates is $y_1+y_2=\frac{-b}{a}$

$=\left(x_1-a\right)y_1$

Product of the ordinates is $y_1{y}_2=\frac{a\left[{\left(x_1+a\right)}^2-y_1^2\right]}{x_1}$.

$\therefore d^2={\left(y_1-y_2\right)}^2={\left(y_1+y_2\right)}^2-4y_1{y}_2$

Now substituting the values we get,

$d^2={\left(x_1-a\right)}^2{y}_1^2-4a{x}_1\left[{\left(x_1+a\right)}^2-y_1^2\right]$

$x_1^2{d}^2=y_1^2\left[{\left(x_1-a\right)}^2+4a{x}_1\right]-4a{x}_1{\left(x_1+a\right)}^2$

$=y_1^2{\left(x_1+a\right)}^2-4a{x}_1{\left(x_1+a\right)}^2$

$={\left(x_1+a\right)}^2\left(y_1-4a{x}_1\right)$

Hence the locus of $\left(x_1,y_1\right)$ is

$x^2{d}^2=\left(y-4a{x}_1\right){\left(x+a\right)}^2$

Hence,

Option $\left(A\right)$ is correct ansewer.