Question

The image of the curve $y^2=4x$ in the line $x+y+2=0$ is

• A. $(x+2{)}^2+4(y-1)=0$
• B. $(x+2{)}^2+4(y+2)=0$
• C. $(x-1{)}^2+4(y-1)=0$
• D. None of these

Answer 1

Answer: (B)

$(x+2{)}^2+4(y+2)=0$

Any point on $y^2=4x$ can be represented by $P(t^2,2t)$ where $t$ is a variable parameter.
If $P^{\prime }(x_1,y_1)$ is the image of $P$ on $x+y+2=0\left(AB\right)$
$P{P}^{\prime }\perp AB$ and the mid-point of $P{P}^{\prime }$ lies on $AB$.
$\therefore \frac{y_1-2t}{x_1-t^2}=1$ and $\frac{x_1+t^2}{2}+\frac{y_1+2t}{2}+2=0$
i.e. $x_1-y_1=t^2-2t$ and $x_1+y_1=-(t^2+2t+4)$
$\therefore$ Adding $2x_1=-4t-4\Rightarrow t=-\frac{(x_1+2)}{2}$
$\therefore$ From $x_1-y_1=\frac{(x_1+2{)}^2}{4}+(x_1+2)$
$\Rightarrow$ The locus of $(x_1,y_1)$ is $(x+2{)}^2+4(y+2)=0$

Hence, option B.